{color{red}(a)} Collecting the data to complete the table:\
begin{center}
begin{tabular}{|c||c|||}
hline
Distance from wall (inches)& Width of field of view (inches) \
hlinehline
$144$&$20.7$ \
hline
$132$&$19.6$ \
hline
$120$&$17.3$ \
hline
$108$&$16.2$ \
hline
$96$&$14.8$ \
hline
$84$&$13.1$ \
hline
$72$&$11.4$ \
hline
$60$&$9.3$ \
hline

end{tabular}
end{center}
vspace{ 3 mm}
{color{red}(b)} plotting the data: the association between the width of the field of view and distance from the wall is positive linear association.\

$text{color{#c34632}(c)}$ We use two points lie on the best fit line to get the $y$ intercept and the slope.

using the points $(72, 11.4)$ and $(108, 16.2)$

Subtitute with $Rightarrow y_2 = 16.2, y_1 = 11.4, x_2= 108$ and $x_1=72$ in the form of slope $color{#c34632}m = dfrac{y_2 – y_1}{x_2 – x_1}$

$Rightarrow m= dfrac{108-72}{16.2-11.4} = dfrac{36}{4.8} Rightarrow boxed{m=dfrac{2}{15}}$

Using the point $color{#4257b2} (72, 11.4)$ to get the $(b)$ by substituting with the value of slope in the equation $y= mx+b$ $Rightarrow 11.4= – dfrac{2}{15}(72)+b Rightarrow 11.4= 9.6+b$

Subtract $color{#4257b2} (9.6)$ from both sides $Rightarrow b= 11.4- 9.6 =1.8$

From gragh the $y$ intercept is around $2$

(a) See the table

(b) positive linear association

(c) $m= dfrac{2}{15}$, $y$ intercept $=1.8$

Using results from the previous task, we will first substitute $1800$ in for $x$ and get the following:

$$
1.7+0.15(1800)=271.7
$$

So, Robbie’s field of view will be $271.7$ in at the south end of the field.

Now, we will substitute $6120$ in for $x$ and get the following:

$$
1.7+0.15cdot6120=919.7
$$

So, Robbie’s field of view will be $919.7$ in at the north end of the field.

$271.7$ in, $919.7$ in

#### (a)

Here, you can draw your own football field and use the previous task in order to solve it.

But, let’s imagine that we have already graphed some football field with some dimensions, and using those dimensions, for example, we get the following area of the field of view:

$$
0.8(6.8+23)cdot115=2741.6
$$

So, $textbf{the area is $2741.6$ square yards.}$

#### (b)

The percent of the field which Robbie be able to see would be the following:

$$
dfrac{2741.6}{115cdot52}=0.4584
$$

So, $textbf{Robbie would be able to see $45.84%$ of the field}$.

#### (c)

In this case, Robbie will see $22.9$ yards or:

$$
dfrac{22.9}{52}=0.44
$$

Or, $44.04%$ of the field.

a) $2741.6$; b) $45.84%$; c) $44.04%$

$textbf{(a)}$ There is a strong positive association between weight and length of paint on the pencil. The weight of pencil increases linearly with the increase in the length of paint on the pencil.

The point $2.3$ is an outlier, which means that the pencil having length of paint approximately 2.3 cm is not following the trend.

$textbf{(b)}$ In Sam’s classroom, there can be some fancy pencil’s with different materials of paint on it which puts some extra weight on it, which are related to a particular special brand and so that these are creating the outlier. Moreover, the length of these special pencil is close to 2.3 cm.

$textbf{(c)}$
Given that the equation of the line of best fit is $w=1.4+0.25l$ where $w$ is the weight of the pencil in grams and l is the length of paint on the pencil.

Slope in the above context represent the change in the weight of pencil per cm change in the length of the paint on the pencil.

(d) For the weight of the pencil with length of paint $11.5$ cm, we need to plug the value $l=11.5$ into the given equation of line of best fit.

$$
begin{align*}
w&=1.4+0.25l\
w&=1.4+0.25(11.5)\
w&=1.4+2.875\
w&=4.275 text{grams}
end{align*}
$$

(e) $y$-intercept in this context represents the weight of the pencil with the length of paint on it equal to zero.

$textbf{(a)}$ There is a strong positive association between weight and length of paint on the pencil. The point $2.3$ is an outlier.

$textbf{(b)}$ In Sam’s classroom, there can be some fancy pencil’s with different materials of paint on it which puts some extra weight on it, which are related to a particular special brand and so that these are creating the outlier.

$textbf{(c)}$Slope in the above context represent the change in the weight of pencil per cm change in the length of the paint on the pencil.

$textbf{(d)}$ 4.275 grams

$textbf{(e)}$ $y$-intercept in this context represents the weight of the pencil with the length of paint on it equal to zero.

$$
{color{#4257b2}text{a)}}
$$

$$
begin{align*}
left(3xy^3right)left(-2yright)& {=}quad : -3xy^3cdot :2ytag{Remove parentheses}\
&=-6xy^3y tag{Multiply the numbers} \
&=-6x:y^{3+1} \
&={color{#c34632}-6xy^4}
end{align*}
$$

$$
{color{#4257b2}text{b)}}
$$

$$
begin{align*}
left(x^3y^2right)left(x^{-2}y^{-2}right)& {=}quad : x^3y^2x^{-2}y^{-2}tag{Remove parentheses}\
&=:x^{3-2}y^{-2}y^2 \
&=xy^0=1cdot 😡 \
&={color{#c34632}x}
end{align*}
$$

$$
color{#c34632} text{ }mathrm{Apply:exponent:rule}:quad :a^bcdot :a^c=a^{b+c}
$$

$$
text{ }
$$

$$
text{ }
$$

$$
{color{#4257b2}text{c)}}
$$

$$
begin{align*}
frac{6x^{-2}}{3x^2}& {=}quad : frac{2x^{-2}}{x^2}tag{Divide the numbers}\
&frac{x^{-2}}{x^2}=frac{1}{x^{2-left(-2right)}}=frac{1}{x^4} tag{Simplify} \
&={color{#c34632}frac{2}{x^4}}
end{align*}
$$

$$
color{#c34632} text{ }mathrm{Apply:exponent:rule}:quad frac{x^a}{x^b}:=:frac{1}{x^{b-a}}
$$

$$
text{ }
$$

$$
text{ }
$$

$$
{color{#4257b2}text{d)}}
$$

$$
begin{align*}
left(-2xright)^{-3}& {=}quad :frac{1}{left(-2xright)^3}\
&=frac{1}{-left(2xright)^3} \
&=frac{1}{-2^3x^3}\
&={color{#c34632}-frac{1}{8x^3}}
end{align*}
$$

$$
color{#c34632}text{ }mathrm{Apply:exponent:rule}:quad left(-aright)^n=-a^n,:mathrm{if:}nmathrm{:is:odd}
$$

$$
color{#c34632} text{ }mathrm{Apply:exponent:rule}:quad left(acdot :bright)^n=a^nb^n
$$

$$
text{ }
$$

$$
color{#4257b2} text{ a) }-6xy^4
$$

$$
color{#4257b2} text{ b) }x
$$

$$
color{#4257b2}text{ c) }frac{2}{x^4}
$$

$$
color{#4257b2} text{ d) }-frac{1}{8x^3}
$$

$$
text{ }
$$

$$
text{ }
$$

The path of pursuit makes the hypotenuse of the right angled triangle created by the distances mentioned in this context.

We can related the length of the sides of a right angled triangle with the pythagoras equation.

$$
text{hypotenuse}=sqrt{text{base}^2+text{height}^2}
$$

In the given context, we need to find the distance between the ball carrier and the defender which is making the base of the said triangle. This can be proceeded as shown below

$$
begin{align*}
text{hypotenuse}&=sqrt{text{base}^2+text{height}^2}\
45&=sqrt{text{base}^2+ 40^2}tag{square both side}\
45^2&=text{base}^2+ 40^2tag{subtract $40^2$ from each side}\
45^2-40^2&=text{base}^2 tag {take square root on each side}\
sqrt{425}&=text{base}\
text{base}&=20.615text{ yards}
end{align*}
$$

Therefore, the distance between the ball carrier and the defender is 21.615 yards.

$bullet$ Given that the distance via freeway is 144 miles and average freeway speed is 75 miles per hour. Time taken to reach San Diego via freeway can be found out as shown below

$$
begin{align*}
text{time}&=dfrac{text{Distance}}{text{Speed}}\
text{time}&=dfrac{144 text{ miles}}{ dfrac{75text{ miles}}{1 text{ hour}}}\
&=(dfrac{144 text{ miles}}{75text{ miles}})cdot (dfrac{1 text{ hour}}{1})\
&=1.92 text{hour}
end{align*}
$$

$bullet$ Now, given that the distance via scenic route is 122 miles and average freeway speed is 65 miles per hour. Time taken to reach San Diego via scenic route can be found out as shown below

$$
begin{align*}
text{time}&=dfrac{text{Distance}}{text{Speed}}\
text{time}&=dfrac{122 text{ miles}}{ dfrac{65text{ miles}}{1 text{ hour}}}\
&=(dfrac{122 text{ miles}}{65text{ miles}})cdot (dfrac{1 text{ hour}}{1})\
&=1.87 text{hour}
end{align*}
$$

We can see that the time taken to reach San Diego via scenic route along the coast is lesser than the time to reach via freeway. Hence, scenic route along the coast is better route.

$textbf{(a)}$
$$
begin{align*}
200x+50&=8000tag{Divide each side by 10 for a simpler equation}\
dfrac{200x}{10}+ dfrac{50}{10}&=dfrac{8000}{10}\
20x+5&=800tag{subtract 5 from each side}\
20x+5-5&=800-5\
20x&=795tag{divide each side by 20}\
dfrac{20x}{20}&=dfrac{795}{20}\
x&=39.75
end{align*}
$$

$textbf{(b)}$For simpler equation multiply each term by least common multiple of 4x and 8, which is 8x
$$
begin{align*}
dfrac{21}{4x}+ dfrac{9}{8}&=2tag{multiple each term by $8x$}\
dfrac{21}{4x}cdot 8x+ dfrac{9}{8}cdot 8x&=2cdot 8x\
42+9x&=16x tag{subtract $9x$ from each side}\
42+9x-9x &=16x-9x\
42&=7xtag{divide each side by 7}\
dfrac{42}{7}&=dfrac{7x}{7}\
x&=6
end{align*}
$$

$textbf{(a)}$ $x=39.75$

$textbf{(b)}$ $x=6$

Lets assume the function is linear and can represented as $f(x)=ax+c$ where, $a$ and $c$ are the constants

We can find the equation by putting the given values in the table and solving for the constants.Then we can check the validity by checking for an additional point. It is satisfies then the equation is correct otherwise we can increase the degree of the equation like $f(x)=ax^2+bx+c$ and so on.
$bullet$ Solving assuming linearity.

$$
begin{align*}
f(x)&=ax+c\
(-7)&=a(2)+c tag{putting $2,-7$ from the table}\
(18)&=a(-3)+c tag{putting $2,-7$ from the table}
end{align*}
$$

Now, we have two equations in two variables, which can be solved as shown below.

$$
begin{align}
-7&=2a+c\
18&=-3a+c
end{align}
$$

Subtract equation $(2)$ from the equation $(1)$ we get

$$
begin{align*}
-7&=2a+c\
-7-18&=2a-(-3a) +c-c\
-25&=5atag{divide each side by 5}\
a&=-5
end{align*}
$$

Now, putting $a=-5$ in equation $(1)$, we get

$$
begin{align*}
-7&=2a+c\
-7&=2(-5)+ctag {add +10 each side}\
-7+10&=-10+10+c\
3&=c\
c&=3
end{align*}
$$

So, the final equation becomes
$$
boxed{f(x)=-5x+3}
$$

Now, checking our final equation for the data point $(0,3)$ for verification.

$$
begin{align*}
f(x)&=-5x+3\
(3)&=-5(0)+3\
3&=3
end{align*}
$$

We can see that our equation is valid.

As our equation is valid for data point $(0,3)$ also, So we don’t need to proceed for the other higher order terms in the equation and our final equation is $$boxed{f(x)=-5x+3}$$
Now, we can solve for outputs by putting the given inputs into this equation. The table is shown below.\\
begin{tabular}{ |p{2cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{3cm}|}
hline
input(x) &2 &10&6&7&-3&0&-10&100&x \
hline
output(f(x)) &-7 &-47&-27&-32&18&3&53&-497&$f(x)=-5x+3$ \
hline
end{tabular}

The equation for the given function in the table is $f(x)=-5x+3$. For the derivation and the table see inside.

There is strong positive and linear association between Net weight of Cereal and the Packaging Cardboard required to pack it. The line of best fit it given by $y=2.2904x-132.1$ where $y$ represents the Net weight of cereal in grams and $x$ is the $text{in}^2$ of cardboard required for the packaging. For this line, slope is equal to $2.2904$ and $y$-intercept is $-132.1$

The predicted amount of Cereal that can be hold by 260 $text{in}^2$ of cardboard packaging can be found out by putting $x=260$ in the equation of line of best fit.

$$
begin{align*}
y&=2.2904x-132.1\
y&=2.2904(260)-132.1\
y&=463.404
end{align*}
$$

Therefore, 260 $text{in}^2$ of cardboard packaging can hold 463.404 grams of Cereal.

There is strong positive and linear association between Net weight of Cereal and the Packaging Cardboard required to pack it. The line of best fit it given by $y=2.2904x-132.1$

260 $text{in}^2$ of cardboard packaging can hold 463.404 grams of Cereal.

$bullet$ Slope in this context represents the amount of Cereal in grams that can be packaged using 1 $text{inch}^2$ of Cardboard .

$bullet$ $y$- intercept represents the amount of Cereal that can be packaged using 0 $text{inch}^2$ of Cardboard. Practically this should be 0 as there is no material and hence no box. But the value given by $y$-intercept is $-132.1$ grams, which doesn’t make sense.

$bullet$ Slope in this context represents the amount of Cereal in grams that can be packaged using 1 $text{inch}^2$ of Cardboard .

$bullet$ $y$- intercept represents the amount of Cereal that can be packaged using 0 $text{inch}^2$ of Cardboard. Practically this should be 0 as there is no material and hence no box. But the value given by $y$-intercept is $-132.1$ which doesn’t make sense.

From the equation of line of best fit in the problem 4-10 we get that 260 $text{in}^2$ box can hold 463.4 grams of cereal but its given that the practically 260 $text{in}^2$ box can hold 355 grams of cereal.

From the given equation in the question we can find the residual as

$$
text{residual}=355-463.4=-108.4
$$

The distance between the actual points (given data) and the line of best fit represents the residual. Hence, on the graph we can represent this distance as residual.

$text{residual}=355-463.4=-108.4$

On the graph, the distance between the actual points (given data) and the line of best fit represents the residual.

The predicted amount of Cereal that can be hold by 471 $text{in}^2$ of cardboard packaging can be found out by putting $x=471$ in the equation of line of best fit.

$$
begin{align*}
y&=2.2904x-132.1\
y&=2.2904(471)-132.1\
y&=1078.7784-132.1\
y&=946.678 text{grams}
end{align*}
$$

Therefore, By the prediction of the line of best fit, 471 $text{in}^2$ of cardboard packaging can hold 946.678 grams of Cereal.

But the from the actual values from the table, 471 $text{in}^2$ of cardboard packaging can hold 1020 grams of Cereal.

Therefore, the residual is given by
$$
text{residual}=1070-946.678=123.322text{ grams}
$$

The residual is indicated on the graph attached below.

$text{residual}=1070-946.678=123.322text{ grams}$
The graph is attached inside with residual indicated in blue colour.

$textbf{(a)}$ The predicted amount of Cereal that can be hold by 600 $text{in}^2$ of cardboard packaging can be found out by putting $x=600$ in the equation of line of best fit.

$$
begin{align*}
y&=2.2904x-132.1\
y&=2.2904(600)-132.1\
y&=1374.24-132.1\
y&=1242.14 text{ grams}
end{align*}
$$

Therefore,it is predicted by the equation that a 600 $text{in}^2$ of cardboard packaging can hold 1242.14 grams of Cereal.

Given that we have 1005 grams of residual for the 600 $text{in}^2$ box. As we know the formula for the residual is
$$
text{residual}=text{actual}-text{predicted}
$$

So, the actual weight can be found out by putting the available values in the equation above

$$
begin{align*}
text{residual}&=text{actual}-text{predicted}\
(1005)&=text{actual}-(1242.14) tag{add 1242.14 to each side}\
1005+1242.14 &= text{actual}-1242.14+1242.14\
1005+1242.14 &= text{actual}\
text{actual}&=2247.14 text{ grams}
end{align*}
$$

$textbf{(b)}$ The reason for such large residual is that the volume of the box made by the cardboard material is increasing cubically so it can accomodate more stuff. The relation is not linear as we were assuming. The linear relation is close for small values and as we can see that the difference becomes greater with large values

$textbf{(a)}$ $text{actual weight}=2247.14 text{ grams}$

$textbf{(b)}$ The reason for such large residual is that the volume of the box made by the cardboard material is increasing cubically so it can accomodate more stuff. The relation is not linear as we were assuming. The linear relation is close for small values and as we can see that the difference becomes greater with large values

$textbf{(a)}$ A cereal with negative residual is means that the actual amount of sugar in Armen’s diet is less than the predicted values by the given equation.
A cereal with negative residual calories is better for Armen’s diet as he should lower the amount of sugar intake with fulfilling his calorie requirement.

$textbf{(b)}$The slope in this problem represents the amount of sugar in grams present in Armen’s diet per unit calorie in his diet.

A negative $y$-intercept in this context represent that if the number of calories in Armen’s diet is taken zero then the amount of sugar in his diet should be negative $(-16.9 text{ grams})$
This makes sense as a diet contains many other things than the sugar which contribute for the calorie. So even if there is not any sugar present in the diet there will be some calories already. So to obtain the 0 calories we need to put a negative value of sugar in the diet. Well, this is a hypothetical condition.

$textbf{(a)}$ A cereal with negative residual is means that the actual amount of sugar in Armen’s diet is less than the predicted values by the given equation.
A cereal with negative residual calories is better for Armen’s diet as he should lower the amount of sugar intake with fulfilling his calorie requirement.

$textbf{(b)}$The slope in this problem represents the amount of sugar in grams present in Armen’s diet per unit calorie in his diet.

A negative $y$-intercept in this context represent that if the number of calories in Armen’s diet is taken zero then the amount of sugar in his diet should be negative $(-16.9 text{ grams})$

Residuals
Residuals are the difference values between the actual values for a situation and the predicted values from the equation devised to represent the situation mathematically.
The formula for residual is given by

$$
text{Residual}=text{Actual}-text{Predicted}
$$

Lesser the value of residual, closer the prediction will be to the actual value. So, a best fit line with small magnitudes or zero magnitude of residuals represents the situation more accurately.

Given that the, home prices in Smallville can be modeled using the equation
$$
p=150+0.041a
$$
where $p$ is the price of the home in thousands of dollars and $a$ is the area of the house in square feet.

Also, Home prices in Fancyville can be modeled using the equation
$$
p=250+0.198a
$$

In the advertisement for a 2800 square foot home was selling for $280,000. We can put the value of$p=280$as$p$is in thousands and$a=2800$ into both the equations then check which of the equation closely satisfies these values and in this manner we can predict the city.

$$
begin{align*}
p&=150+0.041a\
(280)&=150+0.041(2800)\
(280) &=150+114.8\
280&approx264.8
end{align*}
$$

So, Smallville can be the possible city as the price values are very close

$$
begin{align*}
p&=250+0.198a\
(280)&=250+0.198(2800)\
(280) &=150+554.4 \
280&ne 704.4
end{align*}
$$

So, Fancyville cannot be the possible city as there is too much difference in the prices.

To write $2times 10^3 cdot 4times 10^7$ in scientific notation, we can use Jorge’s approach of using Area models in which we can take $2times 10^3$ as one side of rectangle and $4times 10^7$ as other. Then we can find the area by multiplication using the laws of exponent $a^mcdot a^n=a^{m+n}$. So the multiplication can be done as shown below

$$
begin{align*}
&2times 10^3 cdot 4times 10^7 \
&=(2cdot 4)times (10^3cdot 10^7)\
&=8times 10^{(3+7)}\
&=8times 10^{10}
end{align*}
$$

$bullet$ Cadel is right about the answer and this can be explained as shown above.

$bullet$ Lauren is right about the way but her multiplication is incorrect as when we multiply the terms with the exponents then the powers get added as mentioned in the earlier part ($a^mcdot a^n=a^{m+n}$), not that each term gets multiplied like she did in her answer.

So, Cadel is correct and Lauren is incorrect

We can use Jorge’s approach of using Area models in which we can take $2times 10^3$ as one side of rectangle and $4times 10^7$ as other. Then we can find the area by multiplication using the laws of exponent $a^mcdot a^n=a^{m+n}$.

The correct answer is $8times 10^{10}$

We can solve for the values in each part by plugging $f(x)=16$ into the given equations as shown below

$textbf{(a)}$
$$
begin{align*}
&f(x)-2\
&=(16)-2\
&=14
end{align*}
$$

$textbf{(b)}$ We can not calculate $f(2x)$ from the given information as it is not sufficient. The catch here is that $f(2x)ne 2cdot f(x)$

$textbf{(c)}$

$$
begin{align*}
&3f(x)+2\
&=3cdot (16)+2\
&=48+2\
&=50
end{align*}
$$

$textbf{(a)}$ 14

$textbf{(b)}$ We can not calculate $f(2x)$ from the given information as it is not sufficient. The catch here is that $f(2x)ne 2cdot f(x)$

$textbf{(c)}$ 50

Let the age of the van is $x$ years then the age of the truck will be $x+17$ years as per the given conditions. Given that the sum of ages of the vehicles is 41 years.
$Rightarrow$

$$
begin{align*} x+x+17&=41 \
2x+17&=41
end{align*}
$$

Therefore, the situation can be represented by the equation $2x+17=41$ where, $x$ represents the age of blue van in years.

The situation can be represented by the equation $2x+17=41$ where, $x$ represents the age of blue van in years.

$textbf{(a)}$ By applying the transformation $(x,y) rightarrow (-1x,-1y)$ on $Delta$ABC the resultant coordinates are shown below

$$
text{A}(2,2) rightarrow A'(-2,-2)
$$

$$
text{B}(-2,-2) rightarrow B'(2,2)
$$

$$
text{C}(8,-2) rightarrow C'(-8,2)
$$

The possible transformation for this is rotation with centre as mid-point of side AB.

The shape, size and area is preserved in this transformation.

$textbf{(b)}$ By applying the transformation $(x,y) rightarrow (-2x,-2y)$ on $Delta$ABC the resultant coordinates are shown below

$$
text{A}(2,2) rightarrow A”(-4,-4)
$$

$$
text{B}(-2,-2) rightarrow B”(4,4)
$$

$$
text{C}(8,-2) rightarrow C”(-16,4)
$$

The possible transformation for this is rotation with centre as mid-point of side AB and multiplication with a factor 2.

No, this is not a type of rigid transformation.
Only the shape is preserved in this transformation but size and area are different

$textbf{(a)}$ By applying the transformation $(x,y) rightarrow (-1x,-1y)$ on $Delta$ABC the resultant coordinates are shown below

$$
text{A}(2,2) rightarrow A'(-2,-2)
$$

$$
text{B}(-2,-2) rightarrow B'(2,2)
$$

$$
text{C}(8,-2) rightarrow C'(-8,2)
$$

$textbf{(b)}$ By applying the transformation $(x,y) rightarrow (-2x,-2y)$ on $Delta$ABC the resultant coordinates are shown below

$$
text{A}(2,2) rightarrow A”(-4,-4)
$$

$$
text{B}(-2,-2) rightarrow B”(4,4)
$$

$$
text{C}(8,-2) rightarrow C”(-16,4)
$$

For the graph and the explanation see inside.

Using Distributive Property

$$
a(b pm c)=ab pm ac
$$

a)
$$
5x(x-6)
$$

$$
5x cdot x-5x cdot 6=5x^2-30x
$$

b)
$$
-9y(6-3y)
$$

$$
-9y cdot 6+9y cdot 3y=-54y+27y^2=27y^2-54y
$$

$(a) 5x^2-30x$, $(b) 27y^2-54y$

#### (c)

For example, the lowest value can be $150$ and the highest value can be $174.5$ cm.

$textbf{(a)}$ There is strong negative linear association between the number of months of taking supplement and number of days cold lasted. Which means as the number of months of taking supplement increases then the number of cold lasting days decreases.

$textbf{(b)}$ The equation of line of best fit is $y=-1.581x+5.372$. So to find the expected number of days a cold to last for a patient taking supplement for 1.5 months,we can put $x=1.5$ in this equation as shown below

$$
begin{align*}
y&=-1.581x+5.372\
y&=-1.581(1.5)+5.372\
y&=3.0005\
y&approx 3 text{ days}
end{align*}
$$

$textbf{(c)}$ The actual value from the table is, cold lasted 3.3 days for a patient taking supplement for 1.5 months. So the residual is given by

$$
begin{align*}
text{residual}&=text{actual}-text{predicted}\
text{residual}&=3.3-3\
text{residual}&=0.3
end{align*}
$$

The residual for this case is 0.3 days. The positive value of residual here signifies that actually the cold last for 0.3 days longer than the predicted value by the equation of line of best fit.

$textbf{(d)}$ The $y$-intercept in this context represents the number of days cold lasts for a patient who have not taken the supplement (taken for 0 month)

$textbf{(a)}$ There is strong negative linear association between the number of months of taking supplement and number of days cold lasted. Which means as the number of months of taking supplement increases then the number of cold lasting days decreases.

$textbf{(b)}$ 3 days

$textbf{(c)}$ 0.3 days

$textbf{(d)}$ The $y$-intercept in this context represents the number of days cold lasts for a patient who have not taken the supplement (taken for 0 month).

$textbf{(a)}$ The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

So, the equation of a line of slope $-dfrac{2}{3}$ passing through a point $(-12,16)$ is given by

$$
begin{align*}
y-(16)&=-dfrac{2}{3}cdot (x-(-12))\
y-16&=-dfrac{2}{3}cdot x -dfrac{2}{3}cdot 12\
y-16&=-dfrac{2}{3}cdot x-8\
y&=-dfrac{2}{3}cdot x-8+16\
&boxed{y=-dfrac{2}{3}cdot x+8}
end{align*}
$$

$textbf{(b)}$ The $x$-intercept is the value of the variable $x$ when variable $y=0$, so we can find the $x$-intercept by plugging $y=0$ into the equation of line derived above.

$$
begin{align*}
y&=-dfrac{2}{3}cdot x+8\
0&=-dfrac{2}{3}cdot x+8\
-8&=-dfrac{2}{3}cdot x\
-dfrac{3}{2} cdot -8&=-dfrac{3}{2}-dfrac{2}{3}cdot x\
-3cdot -4&=1cdot x\
12&=x\
&boxed{x=12}
end{align*}
$$

$textbf{(a)}$ $y=-dfrac{2}{3}cdot x+8$

$textbf{(b)}$ $x=12$

$textbf{(a)}bullet$ Area as sum of its parts

$$
begin{align*}
text{Area}&=xy+3y+y^2\
&=y^2+xy+3y
end{align*}
$$

$bullet$ Area as product of the dimensions

$$
begin{align*}
text{Area}&=(y)cdot(x+3+y)\
&=ycdot x+ycdot 3+ycdot x y\
&=xy+3y+y^2\
&=y^2+xy+3y
end{align*}
$$

We can see that Area from both the cases are the same hence, we can write it as shown below.

$$
text{Area}=(y)cdot(x+3+y)=y^2+xy+3y
$$

$$
textbf{(b)}
$$

$bullet$ Area as sum of its parts

$$
begin{align*}
text{Area}&=x^2+3x+8x+24\
&=x^2+11x+24
end{align*}
$$

$bullet$ Area as product of the dimensions

$$
begin{align*}
text{Area}&=(x+8)cdot(x+3)\
&=xcdot(x+3)+8cdot (x+3)\
&=x^2+3x +8x+24\
&=x^2+11x+24
end{align*}
$$

We can see that Area from both the cases are the same hence, we can write it as shown below.

$$
text{Area}=(x+8)cdot(x+3)=x^2+11x+24
$$

$textbf{(a)}$ $text{Area}=(y)cdot(x+3+y)=y^2+xy+3y$

$textbf{(b)}$ $text{Area}=(x+8)cdot(x+3)=x^2+11x+24$

Given functions

$$
f(x)=3-|x|
$$

$$
g(x)=3^x+5
$$

Now plugging in values into given functions and calculating:

a) $f(-5)=3-|-5|=3-5=-2$

b) $g(4)=3^4+5=81+5=86$

c) $f(0)=3-0=3$

d) $f(2)=3-|2|=3-2=1$

e) $g(1)=3^1+5=3+5=8$

f) $g(0)=3^0+5=1+5=6$

a) $f(-5)=-2$

b) $g(4)=86$

c) $f(0)=3$

d) $f(2)=1$

e) $g(1)=8$

f) $g(0)=6$

$textbf{(a)}$ For writing the simpler equation, we can remove the fractional terms from the equation, which can be done by multiplying each term by the Least common multiple of the denominators.

$$
begin{align*}
dfrac{2}{7}x+2&=dfrac{3}{10}tag{multiply each term by 70}\
dfrac{2}{7}x cdot 70 +2cdot 70 &=dfrac{3}{10}cdot 70 \
2xcdot dfrac{70}{7} +140&= 3cdot dfrac{70}{10}\
20x+140&=21tag{subtract 140 from each side}\
20x+140-140&=21-140\
20x&=-119tag{divide each side by 20}\
&boxed{x=-dfrac{119}{20}}
end{align*}
$$

Now, checking our solution in the original equation

$$
begin{align*}
dfrac{2}{7}x+2&=dfrac{3}{10}\
dfrac{2}{7}(-dfrac{119}{20})+2&=dfrac{3}{10}\
-dfrac{2}{20}cdot dfrac{119}{7}+ 2cdot dfrac{10}{10}&=dfrac{3}{10}tag{take $dfrac{1}{10}$ common }\
dfrac{1}{10}(-dfrac{119}{7}+20)&=dfrac{3}{10}tag{multiply each side by 10}\
( dfrac{-119+140}{7})&=3\
dfrac{21}{7}&=3\
3&=3
end{align*}
$$

We can see that the solution $x=-dfrac{119}{20}$ is satisfying the original equation, hence verified.

$textbf{(a)}$ For writing the simpler equation, we can remove the fractional terms from the equation, which can be done by multiplying each term by the Least common multiple of the denominators.

$$
begin{align*}
dfrac{9}{50}+dfrac{6x}{25}&=dfrac{3}{10}tag{multiply each term by 50}\
dfrac{9}{50}cdot 50+dfrac{6x}{25}cdot 50&=dfrac{3}{10}cdot 50\
9cdot dfrac{50}{50} +6xcdot dfrac{50}{25}&= 3cdot dfrac{50}{10}\
9+12x&=15tag{subtract 9 from each side}\
9-9+12x&=15-9\
12x&=6tag{divide each side by 6}\
&boxed{x=dfrac{1}{2}}
end{align*}
$$

Now, checking our solution in the original equation

$$
begin{align*}
dfrac{9}{50}+dfrac{6x}{25}&=dfrac{3}{10}\
dfrac{9}{50}+dfrac{6(dfrac{1}{2})}{25}&=dfrac{3}{10}\
dfrac{9}{50}+dfrac{3}{25}&=dfrac{3}{10}\
dfrac{9}{50}+dfrac{6}{50}&=dfrac{3}{10}\
dfrac{9+6}{50}&=dfrac{3}{10}\
dfrac{15}{50}&=dfrac{3}{10}\
dfrac{3}{10}&=dfrac{3}{10}
end{align*}
$$

We can see that the solution $x=dfrac{1}{2}$ is satisfying the original equation $dfrac{9}{50}+dfrac{6x}{25}=dfrac{3}{10}$, hence verified.

$textbf{(a)}$ $x=-dfrac{119}{20}$

$textbf{(b)}$ $x=dfrac{1}{2}$

$textbf{(a)}$ $S+L$ represents the combined volume of small and big espressos.

$textbf{(b)}$ $x+y$ represents the total number of the drinks sold on a given day

$textbf{(c)}$ $xS+yL$ represents the total volume of the drink sold on a given day. because $xS$ is the volume of total number of small drinks on a given day and similarly $yL$ is the volume of total number of large drinks on a given day.

$textbf{(a)}$ $S+L$ represents the combined volume of small and big espressos.

$textbf{(b)}$ $x+y$ represents the total number of the drinks sold on a given day

$textbf{(c)}$ $xS+yL$ represents the total volume of the drink sold on a given day

$textbf{(a)}$ The graph is drawn below with the line of best fit. From the plot we can see the line of best fit approximately passes through the points $(671,185)$ and $(2825,1496)$. So we can write the equation of the line passing through these points. The procedure is explained below.

Slope of a line which passes through a pair or points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$$
text{slope}=dfrac{y_2-y_1}{x_2-x_1}
$$

Now, we have our set of points as $(671,185)$ and $(2825,1496)$ . Let $(x_1,y_1)=(671,185)$ and $(x_2,y_2)=(2825,1496)$, therefore slope of line passing through these point is
$$
begin{align*}
text{slope}&= dfrac{1496-185}{2825-671}\
&=dfrac{1311}{2154}\
&=0.6086
end{align*}
$$

The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Therefore, the equation of a line of slope $0.6086$ passing through a point $(671,185)$ is given by

$$
begin{align*}
y-(185)&=0.6086cdot (x-(671))\
y-185&=0.6086cdot x +0.6086cdot -671\
y-185&=0.6086cdot x -408.3706\
y&=0.6086cdot x -408.3706+185\
y&=0.6086cdot x -222.3706
end{align*}
$$

$textbf{(b)}$ The data point $(3090,2491)$ is an outlier for this data. This is an outlier as it is not following the trend and has a high magnitude of residual. On the scatterplot it is marked with red colour. This data points corresponds to the Sorden, Scottie.

The residual is defined as shown below
$$
text{Residual}=text{actual value}-text{predicted value}
$$

The actual value of point from the table is $2491$ and predicted value can be calculated from equation of the line of best fit by putting $x=3090$ in it.

$$
begin{align*}
y&=0.6086cdot x -222.3706\
y&=0.6086cdot (3090) -222.3706\
y&=1880.574 -222.3706\
y&=1658.2034
end{align*}
$$

Therefore, the residual for the data point $(3090,2491)$ is

$$
text{Residual}=2491-1658.2034=832.7966
$$

$textbf{(c)}$ A positive value of residual means that the actual value of points scored by a player is more than the value predicted by the equation of line of best fit. Therefore, a player will be proud of a more positive value of residual.

$textbf{(d)}$ For calculating the points scored Antonio Kusoc we need to put the value of $x$ equal to 2103 minutes in the equation of line of best fit. This can be solved as shown below

$$
begin{align*}
y&=0.6086cdot x -222.3706\
y&=0.6086cdot (2103) -222.3706\
y&=1279.8858 -222.3706\
y&=1057.5152
end{align*}
$$

Therefore, Antonio Kusoc will score around 1058 points during a season

$textbf{(a)}$ $y=0.6086cdot x -222.3706$

$textbf{(b)}$ $text{Residual}=832.7966$

$textbf{(c)}$ A positive value of residual means that the actual value of points scored by a player is more than the value predicted by the equation of line of best fit. Therefore, a player will be proud of a more positive value of residual.

$textbf{(d)}$ Antonio Kusoc will score around 1058 points during a season

After we collect data and find the best fit line, we can find the sum of the
squares of the residuals using the following formula:

$$
RSS=sum_{i=0}^{n} {(y_i – (alpha+beta x_i))^2}
$$

Where $x,y$ are set of values $alpha,beta$ are constant values and $n$ is set value of count.

$$
RSS=sum_{i=0}^{n} {(y_i – (alpha+beta x_i))^2}
$$

textbf{(c)}\ The residuals are represented into a table shown below, the order is following the player’s order in the given table.\\
begin{tabular}{ |p{2cm}|p{1.5cm}|p{1.5cm}|p{1.5cm}|p{1.5cm}|p{1.5cm}|p{1.5cm}|p{1.5cm}|}
hline
multicolumn{8}{|c|}{Residuals} \
hline
862.117&24.712&-318.864&-203.163&-244.489&-681.994&-48.6155&\
hline
209.5771&46.662&17.37&17.3705&143.792&66.628&-5.303&131.152\
hline
end{tabular}
\\
The sum of the squares of the residuals is 1,502,806.\\
textbf{(d)} For the scores of Antonio Kusoc we need to put $x=2103$ in the equation $y=0.5947x-208.47$.
begin{align*}
y&=0.5947x-208.47\
y&=0.5947(2103)-208.47\
y&=1250.6541-208.47\
y&=1042.1841
end{align*}
Therefore, the predicted points for the Antonio Kusoc from the LSRL is approximately 1042.\\
textbf{(e)} The slope in this context represents the number of points scored by a player per minute.\
The $y$-intercept in this context tells the points scored by a players if he plays for 0 minutes. Practically it should be zero but due to imperfections in the modelling there is some non zero value for $y$-intercept.\
The LSRL model is not reasonable for players that played less than 350 minutes as for those players it is showing the number of points scored as negative which is not possible.\

$textbf{(c)}$ The sum of the squares of the residuals is 1,502,806.

$textbf{(d)}$ The predicted points for the Antonio Kusoc from the LSRL is approximately 1042.

$textbf{(e)}$The LSRL model is not reasonable for players that played less than 350 minutes as for those players it is showing the number of points scored as negative which is not possible.
For more explanation see inside

$textbf{(a)}$ Slope of the line gives the increase in the price of the hamburger per ounce increase in the amount of meat in the patty, which is $0.735 per ounce for the given case.\
Y-intercept gives the cost of hamburger when there is no meat patty in it that means only the cost of the bun. Here, in this case, it’s $0.253 and this does make sense in this situation.\\
textbf{(b)} Let’s find out the cost of a hamburger with the 3-ounce patty in it, by the given LSRL equation by putting a value of$x=3
$$
begin{align*}
y &=0.253 + 0.735times(3)\
y&=0.253+2.205\
y&=$2.458\
y&approx $2.46
end{align*}
But the cost of a hamburger with 3-ounce patty is given $3.20. Hence the residual is
$$
$$(3.20-2.46)=$0.74$ $which signifies that the LSRL equation gives the cost of a hamburger with 3-ounce patty $0.74 lower than the actual available hamburger price in the market. \\
textbf{(c)} 1 pound is equivalent to 16 ounce. Assuming that all the weight is of meat patty by neglecting the weight of the bun, we can find the cost of hamburger with 16-ounce patty by putting$x=16$ in the LSRL equation as shown below.

$$
begin{align*}
y &=0.253 + 0.735times(16)\
y&=0.253+11.76\
y&=$12.013
end{align*}
$$

The cost of 1 pound hamburger is close to $12.013 which is lower than the cost of a hamburger in his friend’s town. Therefore, we can say that the price is reasonable in Charlie’s town.

(a) The slope of the line gives the increase in the price of the hamburger per ounce increase in the amount of meat in the patty.

Y-intercept gives the cost of hamburger when there is no meat patty in it that means only the cost of a bun.

(b) LSRL equation gives the cost of hamburger with 3-ounce patty $0.74 lower than the actual available hamburger price in the market.

(c) The cost of 1 pound hamburger is close to $12.013 which is lower than the cost of a hamburger in his friend’s town. Therefore, we can say that the price is reasonable in Charlie’s town.

The general form of equation of line is written as $y=mx+c$ where, $m$ represents the slope of line and $c$ represents the $y$-intercept of the line.

The equation $2x+3y=6$ can be rearranged and can be written in the general form so that we can compare it with the general form equation of line for the slope and the $y$-intercept.

$$
begin{align*}
2x&+3y=6tag{add -2x to each side}\
2x-2x+3y&=-2x+6\
3y&=-2x+6tag{divide each term by 3}\
dfrac{3y}{3}&=dfrac{-2}{3}x+dfrac{6}{3}\
y&=-dfrac{2}{3}x+2
end{align*}
$$

On comparing the equation $y=-dfrac{2}{3}x+2$ with $y=mx+c$ we can say that

$$
text{Slope}=-dfrac{2}{3}
$$

$$
ytext{-intercept}=2
$$

from drawing graph for the line we need to locate at least 2 points on the graph through which we can draw the line.
The $y$-intercept is where $x=0$, therefore one point on this line is $(0,2)$

For other point, we can put $y=0$ in this line

$$
begin{align*}
2x+3y&=6\
2x+3(0)&=6\
2x&=6tag{divide each side by 2}\
x&=3
end{align*}
$$

Therefore, the other point on the line is $(3,0)$
Now, we can easily mark these points on the graph and make a line passing through these points.

The graph is attached inside.

$$
text{Slope}=-dfrac{2}{3}
$$

$$
ytext{-intercept}=2
$$

We can verify some of the given points directly from the given graph like we can easily interpret that $(-1,0)$ and $(2,2)$ lie on the line as these are the integral points and can be seen easily. But for non integral points and points which are out of the bound of the printed graph we need to find the equation of line and then check if the given points satisfies the resultant equation of the line.

Slope of a line which passes through a pair or points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$$
text{slope}=dfrac{y_2-y_1}{x_2-x_1}
$$

Now, we have known points as $(-1,0)$ and $(2,2)$ . Let $(x_1,y_1)=(-1,0)$ and $(x_2,y_2)=(2,2)$, therefore slope of line passing through these point is
$$
begin{align*}
text{slope}&= dfrac{2-0}{2-(-1)}\
&=dfrac{2}{2+1}\
&=dfrac{2}{3}
end{align*}
$$

The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Therefore, the equation of a line of slope $dfrac{2}{3}$ passing through a point $(-1,0)$ is given by

$$
begin{align*}
y-(0)&=dfrac{2}{3}cdot (x-(-1))\
y&=dfrac{2}{3}cdot x+dfrac{2}{3}cdot 1\
y&=dfrac{2}{3}cdot x +dfrac{2}{3}
end{align*}
$$

Now, checking for the point $(-4,-2)$ if it is a solution or not. We need to put this point into the equation to check if it satisfies the line or not. If the equation holds true then the point is a solution otherwise not.

$$
begin{align*}
y&=dfrac{2}{3}cdot x +dfrac{2}{3}\
(-2)&=dfrac{2}{3}cdot (-4) +dfrac{2}{3}tag{take LCM on right side}\
-2&=dfrac{-8+2}{3}\
-2&=dfrac{-6}{3}\
-2&=-2text{ Always true }
end{align*}
$$

Therefore, point $-4,-2$ also lies on the given line.

Now, checking for the point $(0,1)$ if it is a solution or not.

$$
begin{align*}
y&=dfrac{2}{3}cdot x +dfrac{2}{3}\
(1)&=dfrac{2}{3}cdot (0) +dfrac{2}{3}tag{take LCM on right side}\
1&=0+dfrac{2}{3}\
1&ne dfrac{2}{3}
end{align*}
$$

Therefore, point $(0,1)$ is not a solution to the Martha’s equation.

We can verify some of the given points directly from the given graph like we can easily interpret that $(-1,0)$ and $(2,2)$ lie on the line as these are the integral points and can be seen easily. But for non integral points and points which are out of the bound of the printed graph we need to find the equation of line and then check if the given points satisfies the resultant equation of the line.
$(-4,-2)$ is also a solution to the Martha’s equation. For a detailed analysis see the answer inside.

$$
{color{#4257b2}text{a)}}
$$

$$
begin{align*}
5^0cdot :2^{-3}& {=}quad : 1cdot :2^{-3}\
&=1cdot frac{1}{2^3}=frac{1}{2^3} \
&={color{#c34632}frac{1}{8}}
end{align*}
$$

$$
color{#c34632} text{ } mathrm{Apply:rule}:a^0=1,:ane :0
$$

$$
color{#c34632} text{ }mathrm{Apply:exponent:rule}:quad :a^{-b}=frac{1}{a^b}
$$

$$
{color{#4257b2}text{b)}}
$$

$$
begin{align*}
frac{a^3}{b^{-2}}cdot frac{ab^2}{a^4}& {=}quad : frac{b^2}{a^3}a^3b^2\
&frac{a^3}{b^{-2}}=frac{a^3}{frac{1}{b^2}}=frac{a^3b^2}{1}=a^3b^2 tag{Simplify} \
&frac{ab^2}{a^4}=frac{b^2}{a^3} tag{Simplify}\
&=frac{b^2a^3b^2}{a^3} tag{Multiply fractions} \
&=b^2b^2 tag{Cancel the common factor} \
&={color{#c34632}b^4}
end{align*}
$$

$$
color{#c34632} text{ } mathrm{Apply:exponent:rule}:quad :a^{-b}=frac{1}{a^b}
$$

$$
color{#c34632} text{ } mathrm{Apply:the:fraction:rule}:quad frac{a}{frac{b}{c}}=frac{acdot :c}{b}
$$

$$
{color{#4257b2}text{c)}}
$$

$$
begin{align*}
2.3cdot :10^{-3}cdot :4.2cdot :10^2& {=}quad : 2.3left(2cdot :5right)^{-3}cdot :4.2cdot :10^2tag{Factor integer}\
&=2.3cdot :2^{-3}cdot :5^{-3}cdot :4.2cdot :10^2 \
&=2.3cdot :2^{-3}cdot :5^{-3}cdot :4.2left(2cdot :5right)^2 tag{Factor integer} \
&=2.3cdot :2^{-3}cdot :5^{-3}cdot :4.2cdot :2^2cdot :5^2 \
&=2^{-1}cdot :5^{-3}cdot :5^2cdot :2.3cdot :4.2 \
&=2^{-1}cdot :5^{-1}cdot :2.3cdot :4.2\
&=2.3cdot :4.2cdot frac{1}{2}cdot frac{1}{5}\
&=2.3cdot :4.2cdot frac{1}{10}\
&=9.66cdot frac{1}{10} tag{Multiply the numbers}\
&=frac{9.66}{10}\
&={color{#c34632}0.966=9.66times 10^{-1}}
end{align*}
$$

$$
color{#c34632} text{ }mathrm{Apply:exponent:rule}:quad left(abright)^c=a^cb^c
$$

$$
color{#c34632} text{ } mathrm{Apply:exponent:rule}:quad :a^bcdot :a^c=a^{b+c}
$$

$$
color{#c34632} text{ } mathrm{Apply:exponent:rule}:quad :a^{-1}=frac{1}{a}
$$

$$
text{ }
$$

$$
{color{#4257b2}text{d)}}
$$

$$
begin{align*}
left(3.5cdot :10^3right)^2& {=}quad : 3.5^2left(10^3right)^2\
&=3.5^2cdot :10^6 \
&=10^6cdot :12.25 \
&=1000000cdot :12.25\
&={color{#c34632}12250000=1.225times 10^{7}}
end{align*}
$$

$$
color{#c34632} text{ }mathrm{Apply:exponent:rule}:quad left(acdot :bright)^n=a^nb^n
$$

$$
color{#c34632} text{ }mathrm{Apply:exponent:rule}:quad left(a^bright)^c=a^{bc}
$$

$$
color{#4257b2} text{ a) }frac{1}{8}
$$

$$
color{#4257b2}text{ b) }b^4
$$

$$
color{#4257b2} text{ c) }9.66times 10^{-1}
$$

$$
color{#4257b2} text{ d) }1.225times 10^{7}
$$

$textbf{(a)}$
$$
begin{align*}
25^2&=125^{x+1}\
(5^2)^2&=(5^3)^{x+1} tag{use $(a^m)^n=a^{mcdot n}$}\
5^{2cdot 2}&=5^{3cdot (x+1)}\
5^4&=5^{3x+3}tag{comparing powers}\
4&=3x+3tag{subtract 3 from each side}\
4-3&=3x+3-3\
1&=3x tag{divide each side by 3}\
dfrac{1}{3}&=dfrac{3x}{3} tag{solve and interchange sides}\
x&=dfrac{1}{3}
end{align*}
$$

$textbf{(b)}$
$$
begin{align*}
dfrac{x}{5}+dfrac{x-1}{3}&=2tag{multiply each term by 15(LCM)}\
dfrac{x}{5}cdot 15+dfrac{x-1}{3}cdot 15&=2cdot 15\
xcdot dfrac{15}{5}+ (x-1)cdot dfrac{15}{3}&=30\
3x+5(x-1)&=30\
3x+5x-5&=30tag{add +5 to each side}\
8x-5+5&=30+5\
8x&=35tag{divide each side by 8}\
dfrac{8x}{8}&=dfrac{35}{8}\
x&=dfrac{35}{8}
end{align*}
$$

$textbf{(a)}$ $x=dfrac{1}{3}$

$textbf{(b)}$ $x=dfrac{35}{8}$

$textbf{(a)}$ In my opinion the Association between the Grocery bill and the number of miles should be negative because mothers living close to groceries store are expected to spent more due to the easy access.

$textbf{(b)}$ Given that upper boundary equation is $g=11.27-0.14d$ and lower boundary equation is $g=7.67-0.14d$

The line of best fit should pass through the middle of the given lines. The $y$-intercept should be the average of the average of the given two equations. Therefore, the value of $y$-intercept is $dfrac{11.27+7.67}{2}=9.47$ and slope will be same as the given lines that is $-0.14$.

Therefore, the equation of a line of slope $-0.14$ passing through a point $(0,9.47)$ is given by

$$
begin{align*}
y&=9.47-0.14x
end{align*}
$$

$textbf{(c)}$ Slope in this context represent the average decrease in amount spent on the grocery bill per unit miles away from the church. Comparing the equation’s slope it is $0.14times 1000=$1400$

$textbf{(d)}$ Given that Fabienne lives 8 miles from the church. Her predicted value of grocery bill can be calculated from the equation by putting $x=8$ into the equation.

$$
begin{align*}
y&=9.47-0.14x\
y&=9.47-0.14(9)\
y&=9.47-1.26\
y&=8.21
end{align*}
$$

As the value of spent is in thousand of dollars therefore, Fabienne’s expenditure on the groceries for the year is $8.21times 1000=$8210$

$textbf{(a)}$ In my opinion the Association between the Grocery bill and the number of miles should be negative because mothers living close to groceries store are expected to spent more due to the easy access.

$textbf{(b)}$ $y=9.47-0.14x$

$textbf{(c)}$ Slope in this context represent the average decrease in amount spent on the grocery bill per unit miles away from the church. Comparing the equation’s slope it is $0.14times 1000=$1400$

$textbf{(d)}$ Fabienne’s expenditure on the groceries for the year is $8.21times 1000=$8210$

$textbf{(a)}$
$$
begin{align*}
6-(3+x)&=10\
6-3-x&=10\
3-x&=10tag{subtract 3 from each side}\
3-3-x&=10-3\
-x&=7tag{divide each side by -1}\
dfrac{-x}{1}&=dfrac{7}{-1}\
&boxed{x=-7}
end{align*}
$$

$textbf{(b)}$
$$
begin{align*}
100(x+3)&=200tag{divide each side by 100}\
dfrac{100(x+3)}{100}&=dfrac{200}{100}\
x+3&=2tag{subtract 3 from each side}\
x+3-3&=2-3\
&boxed{x=-1}
end{align*}
$$

$textbf{(c)}$
$$
begin{align*}
dfrac{1}{3}x+4&=x-2tag{subtract $dfrac{1}{3}x$ from each side}\
dfrac{1}{3}x-dfrac{1}{3}x+4&=x-dfrac{1}{3}x-2\
4&=dfrac{3x-x}{3}-2\
4&=dfrac{2x}{3}-2tag{add 2 to the each side}\
4+2&=dfrac{2x}{3}-2+2\
6&=dfrac{2x}{3}tag{multiple each side by $dfrac{3}{2}$}\
6cdot dfrac{3}{2}&=dfrac{2x}{3}cdot dfrac{3}{2}\
9&=xtag{interchange sides}\
&boxed{x=9}
end{align*}
$$

$textbf{(d)}$
$$
begin{align*}
dfrac{4}{5}&=dfrac{x+2}{45}tag{multiply each side by 45}\
dfrac{4}{5} cdot 45&=dfrac{x+2}{45} cdot 45\
4cdot 9&=x+2\
36&=x+2tag{subtract 2 from each side}\
36-2&=x+2-2\
34&=xtag{interchange sides}\
&boxed{x=34}
end{align*}
$$

$textbf{(a)}$ -7

$textbf{(b)}$ -1

$textbf{(c)}$ 9

$textbf{(d)}$ 34

$textbf{(a)}$ The area of square of given side length $a$ is given by the formula
$$
text{Area}=a^2
$$

We are given that area for a square is 240.56 so the side length for this square can be calculated as shown below

$$
begin{align*}
240.56&=a^2tag{take square root on each side}\
sqrt{240.56}&=sqrt{a^2}\
sqrt{240.56}&=atag{interchange sides}\
&boxed{a=15.5099}
end{align*}
$$

But for the same precision as given for area we need to mention the answer upto 2 digits after the decimals. This will make the side length equal to $15.51$ cms.

$textbf{(b)}$ We can find the length of the diagonal by applying the pythagoras theorem as shown below.

Let the diagonal length is represented by $d$ and side lengths are represented by $a$

$Rightarrow$
$$
begin{align*}
d^2&=a^2+a^2\
d^2&=2a^2tag{take square root each side}\
sqrt{d^2}&=sqrt{2a^2}\
d&=sqrt{2}a
Rightarrow\
d&=1.414cdot 15.51\
d&=21.93114\
&boxed{d=21.93}tag{using same precision as for the area}
end{align*}
$$

$textbf{(a)}$ $15.51$

$textbf{(b)}$ $21.93$

$$
{color{#4257b2}text{a)}}
$$

$$
begin{align*}
5xleft(3xright)& {=}quad : 5xcdot :3xtag{Remove parentheses}\
&=15xx \
&={color{#c34632}15x^2}
end{align*}
$$

$$
color{#c34632} text{ }mathrm{Apply:exponent:rule}:quad :a^bcdot :a^c=a^{b+c}
$$

$$
text{ }
$$

$$
{color{#4257b2}text{b)}}
$$

$$
begin{align*}
5x+3x& {=}quad : color{#c34632}8x\
end{align*}
$$

$$
{color{#4257b2}text{c)}}
$$

$$
begin{align*}
6xleft(xright)& {=}quad : 6xxtag{Remove parentheses}\
&={color{#c34632}6x^2}
end{align*}
$$

$$
color{#c34632} text{ }mathrm{Apply:exponent:rule}:quad :a^bcdot :a^c=a^{b+c}
$$

$$
text{ }
$$

$$
{color{#4257b2}text{d)}}
$$

$$
begin{align*}
6x+x& {=}quad : color{#c34632}7x\
end{align*}
$$

$$
color{#4257b2} text{ a) }15x^2
$$

$$
color{#4257b2} text{ b) } 8x
$$

$$
color{#4257b2}text{ c) }6x^2
$$

$$
color{#4257b2}text{ d) } 7x
$$

$$
{color{#4257b2}text{a)}}
$$

$$
begin{align*}
2x=8& {;}quad : frac{2x}{2}=frac{8}{2}tag{Divide both sides by 2}\\
&{color{#c34632}x=4}
end{align*}
$$

$$
{color{#4257b2}text{b)}}
$$

$$
begin{align*}
2x+2=10& {;}quad : 2x+2-2=10-2tag{Subtract 2 from both sides}\
&2x=8tag{Simplify} \
&frac{2x}{2}=frac{8}{2}tag{Divide both sides by 2} \
&{color{#c34632}x=4}
end{align*}
$$

$$
{color{#4257b2}text{c)}}
$$

$$
begin{align*}
6x+2-4x=10& {;}quad : 6x-4x+2=10tag{Group like terms}\
&2x+2=10tag{Add similar elements} \
&2x+2-2=10-2tag{Subtract 2 from both sides} \
&2x=8tag{Simplify} \
&frac{2x}{2}=frac{8}{2}tag{Divide both sides by 2} \
&{color{#c34632}x=4}
end{align*}
$$

$$
{color{#4257b2}text{d)}}
$$

$$
begin{align*}
2left(3x+1right)-4x=10& {;}quad : 2x+2=10tag{Expand}\
&2x+2-2=10-2tag{Subtract 2 from both sides} \
&2x=8tag{Simplify} \
&frac{2x}{2}=frac{8}{2}tag{Divide both sides by 2} \
&{color{#c34632}x=4}
end{align*}
$$

$$
{color{#4257b2}text{e)}}
$$

$$
color{#c34632}text{ x=4 for every equation }
$$

$$
text{ }
$$

$$
text{ }
$$

$$
color{#4257b2} text{ a) }x=4
$$

$$
color{#4257b2} text{ b) }x=4
$$

$$
color{#4257b2} text{ c) }x=4
$$

$$
color{#4257b2} text{ d) }x=4
$$

$$
color{#4257b2} text{ e) x=4 for every equation }
$$

$$
text{ }
$$

Let Ryan was thinking about number $x$. According to given condition we write the equation

$$
boxed{6x-15=x}
$$

$$
begin{align*}
6x-15&=xtag{subtract x from each side}\
6x-x-15&=x-x\
5x-15&=0tag{add 15 to each side}\
5x-15+15&=0+15\
5x&=15tag{divide each side by 5}\
&boxed{x=3}
end{align*}
$$

The equation for the given situation is $6x-15=x$

Ryan was thinking about the number $3$

$textbf{(a)}$ The equation of LSRL is $y=0.134x+1.66$ and the scatterplot is attached below.

$textbf{(b)}$ Yes, I would consider it as the point of outlier because this point has shifted the LSRL significantly and not following with the trend. The new LSRL is $y=0.064x+7.493$ and the scatter plot is attached.

$textbf{(c)}$The impact of outlier is that it shifted the LSRL significantly below and Amy’s predictions for the field of
view be too small. This can be seen by the comparing the above two LSRL lines on the scatterplots and the actual points on the graph.

$textbf{(a)}$ The equation of LSRL is $y=0.134x+1.66$

$textbf{(b)}$ Yes, I would consider it as the point of outlier because this point has shifted the LSRL significantly and not following with the trend. The new LSRL is $y=0.064x+7.493$

$textbf{(c)}$The impact of outlier is that it shifted the LSRL significantly below and Amy’s predictions for the field of
view be too small.

We can notice that an equation of the best fit line is the following:

$$
y=1.36x+7.74
$$

#### (b)

We will substitute $2$ for $x$ in the previous equation and get the predicted how much Gulia’s father should charge:

$$
y=1.36cdot2+7.74
$$

$$
y=10.46
$$

So, Gulia’s father should charge
$$
10.46$.\
$$

a) $y=1.36x+7.74$; b)
$$
10.46$\
$$

$textbf{(a)}$ Scatterplot I $rightarrow$ Residual Plot C

Scatterplot II $rightarrow$ Residual Plot B

Scatterplot III $rightarrow$ Residual Plot A

$textbf{(b)}$ scatterplot III has a linear model that fit the data best as its residual plot is has lesser magnitude That means the LSRL represents the data points more closely than in the other two cases,

$textbf{(a)}$ The LSRL equation is $y=-0.52x+15.209$

$textbf{(b)}$ Seeing the residual plot we can say that linear model is appropriate as the residual values are very small.
The residual plot tells us that data point collected by Giulia are nearly lies on the LSRL and hence shows a good sign of linearity.

$textbf{(a)}$ The LSRL equation is $y=-0.52x+15.209$

$textbf{(b)}$ Seeing the residual plot we can say that linear model is appropriate as the residual values are very small.
The residual plot tells us that data point collected by Giulia are nearly lies on the LSRL and hence shows a good sign of linearity.

$textbf{(a)}$ No, It is not very good linearisation as residual values are significant and for a appropriate linear model the residual values should be very small in magnitude and close to LSRL on the graph.

$textbf{(b)}$ The LSRL is $f = 9.37 + 3.96p$, where $f$ is the number of avocado farms and $p$ is the population in thousands. For estimating the number of farms for the country with 62,900 people we need to put this value of $p=62.9$ into the LSRL equation.

$$
begin{align*}
f&= 9.37 + 3.96cdot (62.9)\
&= 9.37+249.08\
&=258.45\
&=258
end{align*}
$$

Therefore, the predicted number of avocado farms for the county with a population of 62,900 people is 258

$textbf{(c)}$ From the residual plot we can see that for the input 62.9 we have a residual of approximately $-18$, which means actual number of farms is 18 less than the predicted value by LSRL equation.

Therefore, the actual number of avocado farms in the country with 62,900 residents is

$$
258-18=240
$$

$textbf{(a)}$ No, It is not very good linearisation as residual values are significant and for a appropriate linear model the residual values should be very small in magnitude and close to LSRL on the graph.

$textbf{(b)}$ 258

$textbf{(c)}$ 240

$bullet$ Lets choose three random points $(96,14.6)$ , $(84,13.1)$ and $(72,11.4)$. The Scatterplot corresponding to these points is attached below and the correlation coefficients are also shown in the table.

We can see from the Scatterplot and the LSRL that we have a completely linear relationship between two sets because there are no residuals as all the points lie perfectly on LSRL.

Now, lets take another set of random points $(9,9)$ ; $(20,45)$ and $(36,5)$. The Scatterplot corresponding to these points is attached below and the correlation coefficients are also shown in the table.

We can see from the Scatterplot and the LSRL that we have significant magnitude of residuals and therefore this doesn’t show a appropriate linear relationship.

$textbf{(a)}$ The equation for the LSRL is $y=-1.509x+5.3722$

$textbf{(b)}$ For the upper and lower boundary we need to find the highest positive and highest negative residuals. The upper boundary line will have the same slope as the LSRL but it will pass through the point having the highest positive residual and similarly lower boundary line will have the same slope as the LSRL but it will pass through the point having the highest negative residual.

We can see from the plot that the point $(0.5,5)$ is having highest positive residual. So the equation for the upper boundary line can be written by the procedure explained below.

The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Therefore, the equation of a line of slope $-1.5809$ passing through a point $(0.5,5)$ is given by

$$
begin{align*}
y-5&=-1.5809cdot (x-0.5)\
y-5&=-1.5809cdot x -1.5809cdot (-0.5) \
y-5&=-1.5809cdot x+0.79045 \
y&=-1.5809cdot x+0.79045+5 \
&boxed{y=-1.5809cdot x+5.79045}
end{align*}
$$

Similarly, we can find the equation for lower boundary line. We can see from the plot that the point $(1,2.4)$ is having highest negative residual.

Therefore, the equation of a line of slope $-1.5809$ passing through a point $(1,2.4)$ is given by

$$
begin{align*}
y-2.4&=-1.5809cdot (x-1)\
y-2.4&=-1.5809cdot x -1.5809cdot (-1) \
y-2.4&=-1.5809cdot x+1.5809 \
y&=-1.5809cdot x+1.5809+2.4 \
&boxed{y=-1.5809cdot x+3.9809}
end{align*}
$$

$textbf{(c)}$To predict the length of a cold for a person who has taken supplements for three months, we need to find the values of $y$ in both the upper and lower boundary line equations then we take the magnitude of their difference. It will give the interval

For the $y$ values we need to put $x=3$ in the respective equations

$bullet$ For upper boundary equation

$$
begin{align*}
y&=-1.5809cdot x+5.79045\
y=-1.5809cdot 3+5.79045\
y&=0.99618
end{align*}
$$

$bullet$ For lower boundary equation

$$
begin{align*}
y&=-1.5809cdot x+3.9809\
y=-1.5809cdot 3+3.9809\
y&=-4.7427+3.980\
y&=-0.7627
end{align*}
$$

The magnitude of their difference is $|0.99618-(-0.7627)|=1.7588$.

So applying the same precision in the answer as given in table data, The cold will last around 1.8 days

$textbf{(d)}$ The predicted value for cold period for a person who has never taken the supplement will be given by the value of $y$-intercept of LSRL equation which is around $5.4$ days.

$textbf{(e)}$ For this we need to put value of $x$ equal to 6 into the LSRL equation which is as shown below

$$
begin{align*}
y&=-1.5809cdot x+5.3722\
y&=-1.5809cdot 6+5.3722\
y&=-9.4854+5.3722\
y&=-4.1132
end{align*}
$$

We are getting the number of days as negative, which makes no sense. So, it means after few weeks the person becomes immune to cold and for them cold lasts for 0 days.

$textbf{(f)}$ A negative value of residual means that actually cold lasts for lesser than the predicted value of days from the LSRL equation, hence I will prefer a negative residual value in this situation.

$textbf{(a)}$ The equation for the LSRL is $y=-1.509x+5.3722$

$textbf{(b)}$ Upper boundary equation is $y=-1.5809cdot x+5.79045$ and lower boundary equation is $y=-1.5809cdot x+3.9809$

$textbf{(c)}$ According to the upper and lower boundaries for a person who has taken supplements for three months, the colds lasts for around 1.8 days.

$textbf{(d)}$ $5.4$ days

$textbf{(e)}$ We are getting the number of days as negative, which makes no sense. So, it means after few weeks the person becomes immune to cold and for them cold lasts for 0 days.

$textbf{(f)}$ A negative value of residual means that actually cold lasts for lesser than the predicted value of days from the LSRL equation, hence I will prefer a negative residual value in this situation.

$textbf{(a)}$
$$
begin{align*}
3x+2&=10-4(x-1)\
3x+2&=10-4x+4\
3x+2&=-4x+14tag{add 4x to each side}\
3x+4x+2&=-4x+4x+14\
7x+2&=14tag{subtract 2 from each side}\
7x+2-2&=14-2\
7x&=12tag{divide each side by 7}\
dfrac{7x}{7}&=dfrac{12}{7}\
&boxed{x=dfrac{12}{7}}
end{align*}
$$

$textbf{(b)}$
$$
begin{align*}
4(x-1)-2(3x+5)&=-3x+1tag{applying distributive property}\
4x-4 -6x-10&=-3x+1tag{collect the like terms}\
(4x-6x)+(-4-10)&=-3x+1\
-2x-14&=-3x+1tag{add 2x to the each side}\
-2x+2x-14&=-3x+2x+1\
-14&=-x+1tag{subtract 1 from each side}\
-14-1&=-x+1-1\
-15&=-xtag{multiply each side by -1}\
-15cdot -1&=-xcdot -1\
15&=xtag{interchange sides}\
&boxed{x=15}
end{align*}
$$

$textbf{(a)}$ $x=dfrac{12}{7}$

$textbf{(b)}$ $x=15$

For comparing the speeds we need to convert the given speeds into the same units which can be done by changing either of them.

Given that centipede can travel at 30 meters per minute. Lets find out the speed of centipede into centimeters per second. We know that 1 meter $=$100 centimeters and 1 minute is equivalent to 60 seconds.

$Rightarrow$
$$
begin{align*}
&30 dfrac { text {meters}}{text{minute}}\\
&=30 dfrac { text {meters}}{text{minute}} times dfrac{text{100 centimeters}}{text {1 meter}} times dfrac{text{1 minute} }{text{60 seconds}}\\
&= dfrac{30times 100}{60} dfrac{text{ centimeters}}{text{second}}\\
&=50 dfrac{text{ centimeters}}{text{second}}
end{align*}
$$

Now, we have speed of cockroach as 80 centimeters per second and speed of centipede
as 50 centimeters per second. Therefore, we can easily say that cockroach can travel faster.

Given that the tile pattern has 10 tiles in Figure 2 and increases by 2 tiles for each figure. This implies that Figure 1 should have 2 tiles less than Figure 2 which is $10-2=8$ tiles and Figure 0 should have 6 tiles.

$$
begin{align*}
f(0)&=6\
f(1)&=f(0)+2=6+2=8\
f(2)&=f(1)+2= f(0)+2+2= 6+2cdot 2=10\
f(3)&=f(2)+2= f(0)+2+2+2= 6+2cdot 3=12\
f(n)&=f(n-1)+2= f(0)+2cdot n= 6+2cdot n
end{align*}
$$

Therefore, the equation for this pattern is $f(n)=2cdot n + 6$

Now, for finding the number of tiles in the 100th figure we need to put $n=100$ into the above equation.

$$
begin{align*}
f(n)&=2cdot n + 6\
f(n)&=2cdot 100 + 6\
f(n)&=200 + 6\
f(n)&=206
end{align*}
$$

Hence, the 100th figure will have 206 tiles.

The equation for this pattern is $f(n)=2cdot n + 6$
The 100th figure will have 206 tiles in it.

$textbf{a.}$

From the two points, $Delta x=5-(-3)=8$ and $Delta y=-4-2=-6$. So, the slope is:

$$
m=dfrac{Delta y }{Delta x}=dfrac{-6}{8}=color{#c34632}-dfrac{3}{4}
$$

$textbf{b.}$

Draw a line through the points and a slope triangle with them as the endpoints of the right triangle:

The Pythagorean Theorem states that in a right triangle, the sum of the squares of the two side lengths is equal to the square of the hypotenuse length. In this case, the diagonal is the hypotenuse so we write:

$$
begin{align*}
6^2+8^2&=d^2\
36+64&=d^2\
d^2&=100\
d&=sqrt{100}\
d&=color{#c34632}10
end{align*}
$$

a. $-dfrac{3}{4}$

b. $10$

$textbf{a.}$

Use the exponent rule: $a^{m}cdot a^n=a^{m+n}$

$$
4^2cdot 4^5=color{#c34632}4^7
$$

or if asked to evaluate,

$$
=color{#c34632}16384
$$

$textbf{b.}$

Use the exponent rule: $a^0=1$

$$
(5^0)^3=(1)^3
$$

Evaluate:

$$
=color{#c34632}1
$$

$textbf{c.}$

Use the exponent rule: $a^{m}cdot a^n=a^{m+n}$

$$
x^{-5}cdot x^3=x^{-2}
$$

Use the exponent rule: $a^{-n}=dfrac{1}{a^n}$

$$
=color{#c34632}dfrac{1}{x^{2}}color{white}tag{1}
$$

$textbf{d.}$

Use the exponent rules: $(ab)^n=a^nb^n$ and $(a^{m})^n=a^{mn}$

$$
(x^{-1}cdot y^2)^3=x^{-3}cdot y^6
$$

Use the exponent rule: $a^{-n}=dfrac{1}{a^n}$

$$
=color{#c34632}dfrac{y^6}{ x^{3}}
$$

$textbf{e.}$

Group the decimals and group the powers using associative property:

$$
(8times 10^5)cdot (1.6times 10^{-2})=(8cdot 1.6)times (10^{5}cdot 10^{-2})
$$

Use the exponent rule: $a^{m}cdot a^n=a^{m+n}$

$$
=12.8times 10^3
$$

Write in scientific notation. The decimal part must be greater than or equal to 1 and less than 10 so we move 1 place to the left then add 1 to the power of 10:

$$
=color{#c34632}1.28times 10^4
$$

$textbf{f.}$

Group the decimals and group the powers using associative property:

$$
dfrac{4times 10^3}{5times 10^{5}}=dfrac{4}{5}times dfrac{10^{3}}{10^{5}}
$$

Use the exponent rule: $dfrac{a^{m}}{a^n}=a^{m-n}$

$$
=0.8times 10^{-2}
$$

Write in scientific notation. The decimal part must be greater than or equal to 1 and less than 10 so we move 1 place to the right then subtract 1 from the power of 10:

$$
=color{#c34632}8times 10^{-3}
$$

a. $4^7$ or $16384$

b. $1$

c. $dfrac{1}{x^2}$

d. $dfrac{y^6}{ x^{3}}$

e. $1.28times 10^4$

f. $8times 10^{-3}$

$textbf{(a)}$ The scatterplot is attached below and the equation of LSRL is $y=0.1336x+1.6568$

$textbf{(b)}$ Sum of squares of residuals is $0.58811072$ and the table is attached.

$textbf{(a)}$ $y=0.1336x+1.6568$

$textbf{(b)}$ 0.5881

Given that Figure grows linearly and the number of tiles shown in Figure 3 is 5 and also given that Figure 7 has 13 tiles.

We can represent the situation by assuming that $f(0)=a$ represents the number of tiles in Figure 0 and number of tiles in Figure n is represented by $f(n)$. Now, assume that tiles in every next Figure increased by equal number $c$ more tiles than the previous Figure as the growth pattern is linear.

$$
Rightarrow
$$

$$
begin{align*}
f(0)&=a\
f(1)&=a+c\
f(2)&=a+c+c=a+2c\
f(n)&=a+ncdot c
end{align*}
$$

Now, we can solve for $a$ and $c$ by putting the given values into the question. We have $f(3)=5$

$Rightarrow$
$$
begin{align*}
f(n)&=a+ncdot c\
5&=a+3cdot c
end{align*}
$$

Also, we have given $f(7)=13$

$Rightarrow$
$$
begin{align*}
f(n)&=a+ncdot c\
13&=a+7cdot c
end{align*}
$$

Now, solving the above obtained equation, We can get the values of $a$ and $c$.

$$
begin{align}
5&=a+3cdot c\
13&=a+7cdot c
end{align}
$$

On subtracting the equation (1) from the equation (2), we get

$$
begin{align*}
13-5&=a-a+(7-3)cdot c\
8&=4cdot ctag{divide each side by 4}\
2&=c
end{align*}
$$

Now, putting $c=2$ into equation (1) we get $a=-1$, therefore the final equation becomes
$$
boxed{f(n)=2n-1}
$$

$$
f(n)=2n-1
$$

$$
{color{#4257b2}text{a)}}
$$

$$
begin{align*}
6left(2x-5right)=-left(x+4right)& {;}quad : 12x-30=-x-4tag{Expand}\
&12x-30+30=-x-4+30tag{Add 30 to both sides} \\
&12x=-x+26tag{Simplify} \
&12x+x=-x+26+xtag{Add x to both sides } \
&13x=26tag{Simplify} \
&frac{13x}{13}=frac{26}{13}tag{Divide both sides by 13} \
&{color{#c34632}x=2}
end{align*}
$$

$$
{color{#4257b2}text{b)}}
$$

$$
begin{align*}
left(x+1right)left(x-7right)=left(x-1right)left(x+3right)& {;}quad : x^2-6x-7=x^2+2x-3tag{Expand}\
&x^2-6x-7+7=x^2+2x-3+7tag{Add 7 to both sides} \\
&x^2-6x=x^2+2x+4tag{Simplify} \
&x^2-6x-left(x^2+2xright)=x^2+2x+4-left(x^2+2xright)tag{Subtract x2+2x from both sides } \\
&-8x=4tag{Simplify} \
&frac{-8x}{-8}=frac{4}{-8}tag{Divide both sides by -8} \
&{color{#c34632}x=-frac{1}{2}}
end{align*}
$$

$$
color{#4257b2} text{ a) }x=2
$$

$$
color{#4257b2}text{ b) }x=-frac{1}{2}
$$

For finding the value of $g(-5)$ for each function we need to put the value of $x$ equal to $-5$ in each of the respective functions as shown below.

$textbf{(a)}$
$$
begin{align*}
g(x)&=x^3-2\
g(-5)&=(-5)^3-2\
g(-5)&=(-5cdot -5cdot -5) -2\
g(-5)&=-125-2\
&boxed{g(-5)=-127}
end{align*}
$$

$textbf{(b)}$
$$
begin{align*}
g(x)&=7+sqrt{4-x}\
g(-5)&=7+sqrt{4-(-5)}\
g(-5)&=7+sqrt{4+5}\
g(-5)&=7+sqrt{9}\
g(-5)&=7+3\
&boxed{g(-5)=10}
end{align*}
$$

$textbf{(c)}$
$$
begin{align*}
g(x)&=2^x\
g(-5)&=2^{-5}\
g(-5)&=dfrac{1}{2^5}\
&boxed{g(-5) =dfrac{1}{32}}
end{align*}
$$

$textbf{(d)}$
$$
begin{align*}
g(x)&=1.6x-0.476\
g(-5)&=1.6(-5)-0.476\
g(-5)&=-8-0.476\
&boxed{g(-5) =-8.476}
end{align*}
$$

$textbf{(a)}$ $g(-5) =-127$

$textbf{(b)}$ $g(-5) =10$

$textbf{(c)}$ $g(-5) =dfrac{1}{32}$

$textbf{(d)}$ $g(-5) =-8.476$

$textbf{(a)}$ For showing that the given quadrilateral is parallelogram we need to show that opposite sides are parallel to each other. For this we need to find and show that the opposite side has the same slope.
The procedure for finding the slope is explained below.

Slope of a line which passes through a pair or points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$$
text{slope}=dfrac{y_2-y_1}{x_2-x_1}
$$

Now, we need to pick two vertex points. Let the first pair of point are on segment PN which are $(2,-10)$ and $(1,-7)$. Let $(x_1,y_1)=(1,-7)$ and $(x_2,y_2)=(2,-10)$, therefore slope of line passing through these point is
$$
begin{align*}
text{slope}&= dfrac{-10-(-7)}{2-1}\
&=dfrac{-10+7}{1}\
&=-3
end{align*}
$$

Now, we need to pick vertex points on the opposite side that is on segment QM which are $(-3,-8)$ and $(-4,-5)$
Let $(x_1,y_1)=(-3,-8)$ and $(x_2,y_2)=(-4,-5)$, therefore slope of line passing through these point is
$$
begin{align*}
text{slope}&= dfrac{-5-(-8)}{-4-(-3)}\
&=dfrac{-5+8}{-4+3}\
&=dfrac{3}{-1}\
&=-3
end{align*}
$$

Hence, we can see that the sides QM and PM are parallel to each other as these are having same slope.

Similarly, we can prove for the sides QP and MN. As explained below with the attached figure of the quadrilateral MNPQ.

Let the first pair of point are on segment QP which are $(-4,-5)$ and $(1,-7)$. Let $(x_1,y_1)=(-4,-5)$ and $(x_2,y_2)=(1,-7)$, therefore slope of line passing through these point is
$$
begin{align*}
text{slope}&= dfrac{-7-(-5)}{1-(-4)}\
&=dfrac{-7+5}{1+4}\
&=dfrac{-2}{5}\
end{align*}
$$

Now, we need to pick vertex points on the opposite side that is on segment MN which are $(-3,-8)$ and $(2,-10)$
Let $(x_1,y_1)=(-3,-8)$ and $(x_2,y_2)=(2,-10)$, therefore slope of line passing through these point is
$$
begin{align*}
text{slope}&= dfrac{-10-(-8)}{2-(-3)}\
&=dfrac{-10+8}{2+3}\
&=dfrac{-2}{5}\
end{align*}
$$

Hence, we can see that the sides OP and MN are parallel to each other as these are having same slope. Therefore, indeed the given quadrilateral is a parallelogram.

$textbf{(b)}$ On applying the function $(x rightarrow x, y rightarrow -y)$ the new coordinates become

$$
M(-3,-8) rightarrow M'(-3,8)
$$

$$
N(2,-10) rightarrow N'(2,10)
$$

$$
P(1,-7) rightarrow P'(1,7)
$$

$$
Q(-4,-5) rightarrow Q'(-4,5)
$$

The translated figure is attached below

$textbf{(c)}$ We can attain the shape of M’N’P’Q’ by rotating MNPQ 180$text{textdegree}$. but the name of the vertex points will be changed but still it will be rigid transformation as the shape, size and area is preserved.

$textbf{(a)}$ We can find the slope of the individual sides and see that the slopes of the opposite sides are the same. Hence, given quadrilateral is a parallelogram. For complete explanation see inside.

$textbf{(b)}$ On applying the function $(x rightarrow x, y rightarrow -y)$ the new coordinates become

$$
M(-3,-8) rightarrow M'(-3,8)
$$

$$
N(2,-10) rightarrow N'(2,10)
$$

$$
P(1,-7) rightarrow P'(1,7)
$$

$$
Q(-4,-5) rightarrow Q'(-4,5)
$$

$textbf{(c)}$ We can attain the shape of M’N’P’Q’ by rotating MNPQ 180$text{textdegree}$. but the name of the vertex points will be changed but still it will be rigid transformation as the shape, size and area is preserved.

We can rewrite the simpler equivalent of the equation $dfrac{3}{200}+dfrac{x}{50}=dfrac{7}{100}$ by multiplying each term by 200 then we can easily solve for value of $x$ satisfying the given equation.

$$
begin{align*}
dfrac{3}{200}+dfrac{x}{50}&=dfrac{7}{100} tag{multiply each side by 200}\
dfrac{3}{200}cdot 200+dfrac{x}{50}cdot 200&=dfrac{7}{100}cdot 200\
dfrac{200}{200}cdot 3+dfrac{200}{50}cdot x&=dfrac{200}{100}cdot 7\
3+4cdot x&=2cdot 7\
3+4x&=14 tag{subtract 3 from each side}\
3-3+4x&=14-3\
4x&=11tag{divide each side by 4}\
dfrac{4x}{4}&=dfrac{11}{4}\
&boxed {x=dfrac{11}{4}}
end{align*}
$$

$$
x=dfrac{11}{4}
$$

$bullet$ Lets choose three points $(96,14.6)$ , $(84,13.1)$ and $(72,11.4)$ in the first quadrant. The Scatterplot corresponding to these points is attached and the correlation coefficients are also shown in the table.

As we have a completely linear relationship between two sets, that is why we got value of correlation coefficients $=1$.

$bullet$ In this case , There is no linear relationship between two sets as every point has residual. The LSRL we got has negative slope which shows there is somewhat negative relationship and hence we got value of scatterplot $=-0.19614$

$bullet$ In this case , There is approximately linear relationship between two sets. The LSRL we got has negative slope with negligible residual and shows there is negative relationship and that is why we got value of scatterplot $=-0.9943$ which is very close to the $-1$

Lets choose three points $(96,14.6)$ , $(84,13.1)$ and $(72,11.4)$ in the first quadrant.We have a completely linear relationship between two sets, that is why we got value of correlation coefficients $=1$. The scatterplot and table is attached inside for this case and two other cases with different set of points.

The conclusion is that $r=1$ in this case.

#### (b)

On the following picture, there are two random points whose line has a negative slope.

The conclusion is that $r=1$ in this case.

#### (c)

The conclusion is that the location of all possible points that results $r=1$ is on the line which passes through already graphed two points.

#### (d)

In this case, we conclude that $r$ is negative.

#### (e)

For example, if we add point $(3,25)$, we get that in this case $r=-0.1$.

#### (f)

The largest value of $r$ is $1$ and the smallest value is $-1$.

$r$ close to $1$ that values we have are very related. When $r$ is close to $0$, that means that there is a weak linear relationship between the data.

a) $r=1$; b) $r=1$; c) for points on the line; d) negative; e) $(3,25)$;

f) $-1leq rleq1$

A negative value of correlation coefficient represent the inverse relationship between the input and output that means when input is increasing then output should decrease. In other words the LSRL or the line of best fit for this situation has negative slope.

Similarly, A positive value of correlation coefficient represent the positive relationship between the input and output that means when input is increasing then output should also increase . In other words the LSRL or the line of best fit for this situation has positive slope.

Using the above argument we can see that Scatterplots $textbf{(a.)}$ and $textbf{(d.)}$ should have negative value of correlation coefficients.
Scatterplots $textbf{(b.)}$ and $textbf{(c.)}$ should have positive value of correlation coefficients.

Now, A magnitude close to 1 represents more stronger linear relationship between points on the graph and a magnitude value close to 0 represents more scattered points and very weak correlation between the points on the plot.

Using this argument we can now assign the given value of $r$ to the respective scatterplots which is shown below.

$$
begin{align*}
(r=-0.9) rightarrow text{(a.)}\
(r=-0.6) rightarrowtext{(d.)}\
(r=0.1) rightarrow text{(b.)}\
(r=0.6) rightarrow text{(c.)}
end{align*}
$$

$$
begin{align*}
(r=-0.9) rightarrow text{(a.)}\
(r=-0.6) rightarrowtext{(d.)}\
(r=0.1) rightarrow text{(b.)}\
(r=0.6) rightarrow text{(c.)}
end{align*}
$$

$$
text{Correlation Coefficient}, r
$$

A positive value of correlation coefficient represent the positive relationship between the input and output that means when input is increasing then output should also increase and a negative value of correlation coefficient represent the inverse relationship between the input and output that means when input is increasing then output should decrease.

A magnitude of correlation coefficient $r$ close to 1 represents more stronger linear relationship between the data points and a magnitude value close to 0 represents more scattered points and very weak correlation between the data points.

A positive value of correlation coefficient represent the positive relationship between the input and output that means when input is increasing then output should also increase and a negative value of correlation coefficient represent the inverse relationship between the input and output that means when input is increasing then output should decrease.

A magnitude of correlation coefficient $r$ close to 1 represents more stronger linear relationship between the data points and a magnitude value close to 0 represents more scattered points and very weak correlation between the data points.

We can notice that $r=0.9972$.

#### (b)

Here, the form is linear, direction is the slope, it is $0.13$, the strength is $r$ it is strong.

#### (c)

We can conclude that the form is linear, the direction is increasing and the strength is very strong.

a) $r=0.9972$; b) form is linear, direction is increasing, the strength is very strong; c) form is linear, direction is increasing, the strength is very strong.

$textbf{(a)}$ We have selected some random point on the graph. The scatterplot is drawn along with LSRL equation and the value of $r$ for this data is $0.50736$

$textbf{(b)}$ Scatterplot for each of the category is attached below with the mentioned data points.There are a number of possible scatterplots for each of the category and depends on the choice of data points of different persons. These are for example purpose only.

$textbf{(c)}$ We have chosen five data points such that the association becomes strong negative first.

Now, we chose data point $(15,160)$ to drag it and change its value to draw the separate 4 sets of Scatterplot $C_1$ to $C_4$ and observe the effect on the slopes and the correlation coefficients.

From the sactterplots we can observe that the change in the single value can also affect the slope and the correlation coefficient if the change in the value is very big as we can observe that slope in scatterplot $C_1$ and $C_2$ is negative while it changes to positive in scatterplot $C_3$ and $C_4$.

Similarly, the correlation coefficient becomes smaller and smaller as that point grew as outlier.

We can choose any set of data points for making the scatterplots. For positive association the value of $y$ should increase with the increase in the value of $x$ and for negative association the value of $y$ should decrease with the increase in the value of $x$.
The scatterplots for each part is attached inside and for explicit explanation of each part see inside.

$textbf{(b)}$ Equation of LSRL is $y=4.2151x -6.963$ , where $y$ is the length of pencil and $x$ is the weight

$textbf{(c)}$ Residual plot is attached below.

$textbf{(d)}$ Yes, it seem appropriate to use linear model to make predictions about the weight of pencil as the residual’s magnitude are small except for one or two cases.

$textbf{(e)}$ Using the LSRL equation we can find the length of the pencil of given weight 6g cm.
$$
begin{align*}
y&=4.2151cdot(6) -6.963\
&=25.2906-.963\
&= 18.3276 text{cm}\
end{align*}
$$

Given that the original pencil had 16.8 cm of paint hence the residual is
$$
16.8-18.3276=-1.5276
$$

Now, using the precision of the given data answer should be given in one decimal place. Therefore residual $=-1.5$

$textbf{(f)}$ positive residual means that the estimated answer given by the LSRL equation is lower than the actual value.

$textbf{(b)}$ Equation of LSRL is $y=4.2151x -6.963$ , where $y$ is the length of pencil and $x$ is the weight

$textbf{(d)}$ Yes, it seem appropriate to use linear model to make predictions about the weight of pencil as the residual’s magnitude are small except for one or two cases.

$textbf{(e)}$ -1.5

$textbf{(f)}$ positive residual means that the estimated answer given by the LSRL equation is lower than the actual value.

$$
{color{#4257b2}text{a)}}
$$

$$
begin{align*}
frac{11}{15}+left(-frac{3}{4}right)& {=}quad : frac{11}{15}-frac{3}{4}tag{Remove parentheses}\
&=frac{11cdot :4}{60}-frac{3cdot :15}{60} \
&=frac{4cdot :11-3cdot :15}{60} \
&=frac{-1}{60} \
&={color{#c34632}-frac{1}{60}}
end{align*}
$$

$$
color{#c34632} text{ }mathrm{Apply:the:fraction:rule}:quad frac{-a}{b}=-frac{a}{b}
$$

$$
text{ }
$$

$$
text{ }
$$

$$
{color{#4257b2}text{b)}}
$$

$$
begin{align*}
-5frac{2}{9}-2frac{1}{3}& {=}quad : -frac{47}{9}-frac{7}{3}\
&-5frac{2}{9}=frac{5cdot 9+2}{9}=frac{47}{9} tag{Simplify}\
&2frac{1}{3}=frac{2cdot 3+1}{3}=frac{7}{3} tag{Simplify}\
&=-frac{47}{9}-frac{7cdot :3}{9} \
&=frac{-47-3cdot :7}{9}=frac{-68}{9}=-frac{68}{9} \
&={color{#c34632}-7frac{5}{9}}
end{align*}
$$

$$
color{#c34632} text{ } mathrm{Apply:the:fraction:rule}:quad frac{-a}{b}=-frac{a}{b}
$$

$$
text{ }
$$

$$
text{ }
$$

$$
{color{#4257b2}text{c)}}
$$

$$
begin{align*}
-frac{7}{8}+frac{2}{3}-left(-frac{1}{4}right)& {=}quad : -frac{7}{8}+frac{2}{3}+frac{1}{4}\
&=-frac{7cdot :3}{24}+frac{2cdot :8}{24}+frac{1cdot :6}{24} \
&=frac{-3cdot :7+2cdot :8+1cdot :6}{24} \
&={color{#c34632}frac{1}{24}}
end{align*}
$$

$$
color{#c34632} text{ }mathrm{Apply:rule}:-left(-aright)=a
$$

$$
text{ }
$$

$$
text{ }
$$

$$
{color{#4257b2}text{d)}}
$$

$$
begin{align*}
-3div frac{1}{4}& {=}quad : =-frac{3cdot :4}{1}\
&=-frac{12}{1} \
&={color{#c34632}-12}
end{align*}
$$

$$
color{#c34632} text{ } mathrm{Apply:the:fraction:rule}:quad frac{a}{frac{b}{c}}=frac{acdot :c}{b}
$$

$$
color{#c34632} text{ }mathrm{Apply:rule}:frac{a}{1}=a
$$

$$
text{ }
$$

$$
color{#4257b2}text{ a) }-frac{1}{60}
$$

$$
color{#4257b2} text{ b) }-7frac{5}{9}
$$

$$
color{#4257b2}text{ c) } frac{1}{24}
$$

$$
color{#4257b2} text{ d) }-12
$$

The models are attached above, given dimensions and areas are marked in blue colour and the unknowns are marked in red colour

$textbf{(a)}$ $bullet$ The dimensions for the figure is as follows

$$
text{length}=y-4
$$

$$
text{width}=3y
$$

$bullet$ Area as sum of its parts

$$
begin{align*}
text{Area}&=3y^2-12y
end{align*}
$$

$bullet$ Area as product of the dimensions

$$
begin{align*}
text{Area}&=3ycdot(y-4)\
&=3y(y)+3y(-4)\
&=3y^2-12y
end{align*}
$$

$textbf{(b)}$ $bullet$ The dimensions for the figure is as follows

$$
text{length}=y-4
$$

$$
text{width}=3y+5
$$

$bullet$ Area as sum of its parts

$$
begin{align*}
text{Area}&=3y^2-12y+5y-20\
&=3y^2-7y-20
end{align*}
$$

$bullet$ Area as product of the dimensions

$$
begin{align*}
text{Area}&=(y-4)cdot(3y+5)\
&=ycdot(3y+5)-4cdot (3y+5)\
&=y(3y)+y(5) -4(3y)-4(5)\
&=3y^2+5y-12y-20\
&=3y^2-7y-20
end{align*}
$$

$textbf{(a)}$
$$
text{Dimensions (length, width)}=(y-4,3y)
$$

$$
text{Area}=3y^2-12y
$$

$textbf{(b)}$
$$
text{Dimensions (length, width)}=(y-4,3y+5)
$$

$$
text{Area}=3y^2-7y-20
$$

For determining slope and y-intercept on any given line, we need to rearrange it into the general form of equation of line which is given by

$$
y=mcdot x+c
$$
Where $m$ represents the slope of line and $c$ represents the y-intercept.

$textbf{(a)}$ $2x+7y= 14$ can be written in the general form as shown below

$$
begin{align*}
2x&+7y=14\
7y&=-2x+14\
y&=dfrac{-2}{7}+2
end{align*}
$$

Therefore, Slope of the above line is $dfrac{-2}{7}$ and y-intercept is 2

$textbf{(b)}$ Given equation of line is $y=6-dfrac{x}{3}$ which can be written in the general form as

$$
y=-dfrac{1}{3} x+6
$$

Therefore, Slope of the above line is $-dfrac{1}{3}$ and y-intercept is 6

$textbf{(c)}$ $y=dfrac{10x-2}{2}$ can be simplified and rewritten as
$$
y=5x-1
$$

Therefore, Slope of the above line is $5$ and y-intercept is -1

$textbf{(d)}$ $y=3x$ is already in general form with $c=0$

Therefore, Slope of the above line is $3$ and y-intercept is 0

$textbf{(a)}$ Slope $=dfrac{-2}{7}$ and y-intercept $=2$

$textbf{(b)}$ Slope $=dfrac{-1}{3}$ and y-intercept $=6$

$textbf{(c)}$ Slope $=5$ and y-intercept $=-1$

$textbf{(d)}$ Slope $=3$ and y-intercept $=0$

Given that the local deli sells 4-inch sub sandwiches for $2.95 which means cost per inch of the sandwich is$ $dfrac{$2.95}{4}=$0.7375$ $For the cost of 50 inches long sandwich, we need to multiply 50 inches with the cost per inch as shown below\$Rightarrow
$$
begin{align*}
text{Cost}&=0.7375 dfrac{ $}{ text{ inch}} times 50 text{ inch}\
&=(0.7375times 50) dfrac{$}{ text{inch}} times text{ inch}\
&=$36.875
end{align*}
Therefore, the local deli should charge $36.875 for a 50 inches long sub sandwich.\\
Now, we need to find the unit rate in dollars per foot. We know that 1 foot equals 12 inches. We have cost per inch as $0.7375, so cost of 12 inches sub sandwich can be found out by multiplying it the cost per inch with 12.\
$$
Rightarrow$text{Cost per foot} = 0.7375times 12 = $8.85$$

Local deli should charge $36.875 for a 50 inches long sub sandwich.\
Cost per foot$=$8.85$

For solving this kind of questions, we need to make the same base on the both side so that we can equate the exponents.

$$
textbf{(a)}
$$

$$
begin{align*}
8^x&=2^6\
(2^3)^x&=2^6 tag{use $(a^m)^n=a^{mcdot n}$}\
2^{2cdot x}&=2^6\
2^{2x}&=2^6\
Rightarrow\
2x&=6tag{divide each side by 2}\
&boxed{x=3}
end{align*}
$$

$$
textbf{(b)}
$$

$$
begin{align*}
4^{2x}&=(dfrac{1}{2})^{x+5}\
(2^2)^{2x}&=(dfrac{1}{2})^{x+5} tag{use $(a^m)^n=a^{mcdot n}$}\
2^{4x}&=(dfrac{1}{2})^{x+5}tag{use $a^{-m}=dfrac{1}{a^m}$}\
Rightarrow\
2^{4x}&=2^{-(x+5)} tag{equating the powers}\
4x&=-(x+5)\
4x&=-x-5tag{add $x$ to each side}\
4x+x&=-x+x-5\
5x&=-5tag{divide each side by 5}\
dfrac{5x}{5}&=dfrac{-5}{5}\
&boxed{x=-1}
end{align*}
$$

$textbf{(a)}$ $x=3$

$textbf{(b)}$ $x=-1$

$textbf{(a)}$ There is a positive association between the amount of damage and the number of firefighters at fire. Amount of damage is increased with the number of firefighters at the incident.

$textbf{(b)}$ The decision taken by the mayer is wrong. The number of firefighters at the scene is dependent on how big the fire is and hence the amount of damage. It does not signify that the more number of firefighter cause the damage to the scene. The presence of number of firefighter at the place of fire is secondary result of a big fire. The mayor misunderstood the conveyed information.

$textbf{(a)}$ There is a positive association between the amount of damage and the number of firefighters at fire. Amount of damage is increased with the number of firefighters at the incident.

$textbf{(b)}$ The decision taken by the mayer is wrong. The number of firefighters at the scene is dependent on how big the fire is and hence the amount of damage. It does not signify that the more number of firefighter cause the damage to the scene. The presence of number of firefighter at the place of fire is secondary result of a big fire. The mayor misunderstood the conveyed information.

$textbf{(a)}$ I do not completely agree with the newspaper. If I only eat spinach without taking a complete healthy diet then it will not help me to grow stronger muscles and increase strength
The reason for this association is that there is strong chance that the people adding spinach in their diet are already eating a healthy diet and more concerned with their health which is the reason for their muscles and increased strength.

$textbf{(b)}$ The reason for this association is that spinach has some benefits and considered as one of the healthy food.The diet and health conscious people are more inclined to add spinach in their diet. They are already living a healthy life and spinach is just a single part of their diet.

$textbf{(a)}$ I do not completely agree with the newspaper.The reason for this association is that there is strong chance that the people adding spinach in their diet are already eating a healthy diet and more concerned with their health which is the reason for their muscles and increased strength.

$textbf{(b)}$ The reason for this association is that spinach has some benefits and considered as one of the healthy food.The diet and health conscious people are more inclined to add spinach in their diet. They are already living a healthy life and spinach is just a single part of their diet.

POSSIBLE ANSWERS:

$textbf{a.}$ People may be health conscious so they more likely include calcium intake and be more healthy.

$textbf{b.}$ Busy work schedules or study schedules may be a factor as to why family time declines.

$textbf{c.}$ It is more likely the other way around: People who are depressed eat chocolates to feel good. Since eating too much chocolates can result in weight gain, it might be the reason why people who eat them are depressed.

$textbf{d.}$ Often times, kids are spanked due to misbehavior. This may result in such kids having a hard time in school resulting in lower IQs.

$textbf{e.}$ People may not afford health insurance due to poverty.

$textbf{(a)}$ From the scatterplot below we can say that there is positive and linear association between the wage and the years of experience.

$textbf{(b)}$ From the residual plot we can see that the magnitude of residuals is very small, which tells that the actual points lies very close to LSRL. Therefore, we can say that linear model is appropriate.

$textbf{(c)}$ Correlation coefficient in this case is 0.9984 which is very close to 1. This signifies that there is very strong and positive correlation between hourly wages and work experience which means with increase in the experience the daily wage of a person increases definitely. Correlation coefficient can be calculated using excel or calculator.

$textbf{(a)}$ From the scatterplot below we can say that there is positive and linear association between the wage and the years of experience.

$textbf{(b)}$ Yes, linear model is appropriate as magnitudes of residuals are very small.

$textbf{(c)}$ Correlation coefficient in this case is 0.9984 which is very close to 1. This signifies that there is very strong and positive correlation between hourly wages and work experience which means with increase in the experience the daily wage of a person increases definitely.

$textbf{(a)}$ The slope in this context represents the number of decrease in the attendance (in 1000s) per unit increase in the temperature in $text{textdegree}$ F.

$textbf{(b)}$ The formula for calculating residual is
$$
text{residual}=text{actual}-text{predicted}
$$
. Therefore, the residual for the attendance when temperature were in the 80s for her model is positive which means the value of actual attendance is lower than the predicted attendance. hence, the predictions made by Marissa’s model to is too low when temperature were in 80s.

$textbf{(c)}$ First we need to calculate the predicted value of attendance via Marissa model by putting $t=95$ into the LSRL equation.

$$
begin{align*}
a&=-14+0.41t\
a&=-14+0.41(95)\
a&=-14+38.95\
a&=24.95
end{align*}
$$

As attendance is in 1000s, so we need to multiply it by 1000 for the final predicted value of attendance.

$Rightarrow$ predicted value $=24.95times1000=24950$
Now, from the residual plot, we can see that the residual for attendance when the temperature is 95 is given as $-8times 1000=-8000$. So we can calculate the actual value of attendance by applying the formula
$$
text{residual}=text{actual}-text{predicted}
$$

$Rightarrow$
$$
begin{align*}
text{residual}=text{actual}-text{predicted}\
-8000&=text{actual}-24950\
-8000+24950&=text{actual}\
text{actual}&=16950
end{align*}
$$

Therefore, the actual attendance on the day when the temperature was $text{textdegree}$ 95 is close to 16950.

$textbf{(d)}$ We just need to calculate the value of attendance for the both upper boundary and lower boundary line for the $t=80$. It will give the upper and lower bound respectively.
$bullet$ Calculating upper bound

$$
begin{align*}
a&=-7+0.41t\
a&=-7+0.41(80)\
a&=-7+32.8\
a&=25.8tag{multiply by 1000, for final attendance}\
Rightarrow
a&=25800
end{align*}
$$

$bullet$ Calculating lower bound

$$
begin{align*}
a&=-21+0.41t\
a&=-21+0.41(80)\
a&=-21+32.8\
a&=11.8tag{multiply by 1000, for final attendance}\
Rightarrow
a&=11800
end{align*}
$$

$textbf{(e)}$ The residual values for Marissa’s model is very high. For most of the predicted values her model has a residual whether positive or negative. A better model is such that at least it should be correct of for some of the part without a residual of a very small value of residual. That is why her model does not seems reliable for making predictions.

$textbf{(a)}$ The slope in this context represents the number of decrease in the attendance (in 1000s) per unit increase in the temperature in $text{textdegree}$ F.

$textbf{(b)}$ The predictions made by Marissa’s model to is too low when temperature were in 80s.

$textbf{(c)}$The actual attendance on the day when the temperature was $text{textdegree}$ 95 is close to 16950.

$textbf{(d)}$ 11800 and 25800

$textbf{(e)}$ The residual values for Marissa’s model is very high. For most of the predicted values her model has a residual whether positive or negative. A better model is such that at least it should be correct of for some of the part without a residual of a very small value of residual. That is why her model does not seems reliable for making predictions.

$textbf{a.}$

Use the exponent rule: $a^{-n}=dfrac{1}{a^n}$

$$
2^3cdot 5^{-2}=2^3cdot dfrac{1}{5^{2}}
$$

Multiply:

$$
= dfrac{2^3}{5^{2}}
$$

Evaluate:

$$
= color{#c34632}dfrac{8}{25}
$$

$textbf{b.}$

Use the exponent rules: $(ab)^n=a^nb^n$ and $(a^{m})^n=a^{mn}$

$$
(xy^2)^3cdot (x^{-2})=x^3y^6cdot (x^{-2})
$$

Use the exponent rule: $a^{m}cdot a^n=a^{m+n}$

$$
=color{#c34632}xy^6color{white}tag{1}
$$

$textbf{c.}$

Group the decimals and group the powers using associative property:

$$
3times 10^3cdot 4times 10^5=(3cdot 4)times (10^{3}cdot 10^{5})
$$

Use the exponent rule: $a^{m}cdot a^n=a^{m+n}$

$$
=12times 10^8
$$

Write in scientific notation. The decimal part must be greater than or equal to 1 and less than 10 so we move 1 place to the left then add 1 to the power of 10:

$$
=color{#c34632}1.2times 10^9
$$

$textbf{d.}$

Group the decimals and group the powers using associative property:

$$
dfrac{4times 10^2}{5times 10^{-2}}=dfrac{4}{5}times dfrac{10^{2}}{10^{-2}}
$$

Use the exponent rule: $dfrac{a^{m}}{a^n}=a^{m-n}$

$$
=0.8times 10^4
$$

Write in scientific notation. The decimal part must be greater than or equal to 1 and less than 10 so we move 1 place to the right then subtract 1 from the power of 10:

$$
=color{#c34632}8times 10^3
$$

a. $dfrac{8}{25}$

b. $xy^6$

c. $1.2times 10^9$

d. $8times 10^3$

Recall: The domain is the set of all possible $x$-values of the function. The range is the set of all possible $y$-values of the function.

$textbf{a.}$

The solid points at the end of the segment mean that those points are part of the domain and range. Horizontally, the graph is from $-2$ to 2 so the domain is:

$$
color{#c34632}-2leq xleq 2
$$

Vertically, the graph is from $-3$ to 2 so the range is:

$$
color{#c34632}-3leq yleq 2
$$

$textbf{b.}$

A vertical line suggests that there is only a single value for $x$. So, the domain is:

$$
color{#c34632}x=2
$$

Vertically, the graph extends below and above so the range is:

$$
color{#c34632}text{all real numbers}
$$

$textbf{c.}$

The solid point at the left end of the graph means that the point is part of the domain. Horizontally, the graph is from $-2$ extending to the right so the domain is:

$$
color{#c34632} xgeq -2
$$

Vertically, the graph extends down to the right and up to the right so the range is:

$$
color{#c34632}text{all real numbers}
$$

$textbf{d.}$

Only graph (a) is a function because for every $x$-value, there is only one $y$-value corresponding to it.

a. D: $-2leq xleq 2$ ; R: $-3leq yleq 2$

b. D: $x=2$ ; R: all real numbers

c. D: $xgeq -2$ ; R: all real numbers

d. Graph a

Given that the function is $f(x)=4^x+1$, for calculating the values by putting respective inputs we need to plug the value of $x$ into the mentioned equation and solve for the value of $f(x)$.

$textbf{(a)}$
$$
begin{align*}
f(x)&=4^x+1\
f(0)&=4^0+1tag{use $a^0=1$}\
f(0)&=1+1\
&boxed{f(0)=2}
end{align*}
$$

$textbf{(b)}$ The value of $sqrt[3]{-27}=sqrt[3]{-3cdot -3cdot -3}=-3$

Therefore, we need to find $f(-3)$ which can be proceeded as shown below.

$$
begin{align*}
f(x)&=4^x+1\
f(sqrt[3]{-27})&=4^{-3}+1tag{use $a^{-m}=dfrac{1}{a^m}$}\
f(sqrt[3]{-27})&=dfrac{1}{4^{3}}+1\
f(sqrt[3]{-27})&=dfrac{1}{64}+1\
f(sqrt[3]{-27})&=dfrac{1+64}{64}tag{by taking LCM }\
&boxed{f(sqrt[3]{-27})=dfrac{65}{64}}
end{align*}
$$

$textbf{(c)}$ To find the value of $x$ if $f(x)=17$

$$
begin{align*}
f(x)&=4^x+1\
17&=4^x+1tag{subtract 1 from each side}\
17-1&=4^x+1-1\
16&=4^x\
4^2&=4^xtag{compare powers as base is same}\
2&=x\
&boxed{x=2}
end{align*}
$$

$textbf{(a)}$ $2$

$textbf{(b)}$ $dfrac{65}{64}$

$textbf{(c)}$ $2$

Slope of a line which passes through a pair or points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$$
text{slope}=dfrac{y_2-y_1}{x_2-x_1}
$$

Now, we have given points as $(-3,1)$ and $(9,7)$. Let $(x_1,y_1)=(-3,1)$ and $(x_2,y_2)=(9,7)$, therefore slope of line passing through these point is
$$
begin{align*}
text{slope}&= dfrac{7-1}{9-(-3)}\
&=dfrac{6}{12}\
&=dfrac{1}{2}
end{align*}
$$

The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Therefore, the equation of a line of slope $dfrac{1}{2}$ passing through a point $(9,7)$ is given by

$$
begin{align*}
y-(7)&=dfrac{1}{2}cdot (x-9)\
y-7&=dfrac{1}{2}cdot x + dfrac{1}{2}cdot -9\
y&=dfrac{1}{2}cdot x +dfrac{-9}{2}+7\
y&=dfrac{1}{2}cdot x +dfrac{-9+14}{2}\
&boxed{y=dfrac{1}{2}cdot x +dfrac{5}{2}}
end{align*}
$$

$$
y=dfrac{1}{2}cdot x +dfrac{5}{2}
$$

$text{R}^2$ is called coefficient of determination which can be obtained by squaring the coefficient of correlation.

Unlike coefficient of correlation($r$), coefficient of determination($text{R}^2$) varies between 0 to 1 as it is a square of a value between -1 to 1. Coefficient of determination indicates the extent to which the dependent variable is predictable.

If $text{R}^2=0$ then it means that dependent variable cannot be predicted from the independent variable. And $text{R}^2=1$ means that the dependent variable can be exactly predicted from the independent variable without any error.

For example $text{R}^2=0.25$ means that 25 percent of the variance in output $y$ is
predictable from the input $x$.

$textbf{(a)}$ The electronically calculated (from excel) value of $text{R}^2$ for the given data is equal to $0.8616approx0.86$ which means that the 86 percent variance in height can be predicted by the shoe size.

$textbf{(b)}$ The other factor affecting the height of a person can be
$bullet$ Genes from the family

$bullet$ Sex of the person

$bullet$ Age

$bullet$ Body weight

$textbf{(a)}$ The electronically calculated (from excel) value of $text{R}^2$ for the given data is equal to $0.8616approx0.86$ which means that the 86 percent variance in height can be predicted by the shoe size.

$textbf{(b)}$ The other factor affecting the height of a person can be
$bullet$ Genes from the family

$bullet$ Sex of the person

$bullet$ Age

$bullet$ Body weight

$textbf{(a)}$ The correlation coefficient for the given data is 1 (calculated via excel). This can also verified by graphing the points on the axis and we can see that a straight line passes through all these points which means there is perfect strong positive linear association between the points.
This is unusual as foot size cannot be perfectly linearly related with the height as there are many other deciding factors.

Hence, the association between the height and the shoe size is very strong, positive and linear , which means with the increase in height the shoe size is linear.

$textbf{(b)}$ The height of students in her class varies between 153 cm to 170 cm with average height of student 62.8 cm.
The sex can be also a deciding factor, students engaged in active sports like basket ball are taller than the average.

$textbf{(a)}$ The correlation coefficient for the given data is 1 (calculated via excel). This can also verified by graphing the points on the axis and we can see that a straight line passes through all these points which means there is perfect strong positive linear association between the points.

$textbf{(b)}$ The height of students in her class varies between 153 cm to 170 cm with average height of student 62.8 cm.
The sex can be also a deciding factor, students engaged in active sports like basket ball are taller than the average.

$textbf{(a)}$ The data is almost completely scattered and the value of correlation coefficient for this data should be close to 0.

As the value of correlation coefficient is close to zero, which means its square is also close to zero which means that coefficient of determination $text(R)^2$ is close to 0 and hence a very less percentage of variability in girls height can be predicted by their score in the Chapter 5 Test.

$textbf{(b)}$ The line of best best should be a horizontal line dividing the completely scattered points in equal halves so that the residual magnitudes and the number of points above and below to the line of best fit should be same.

As it is given that the average height of girls in her cass is 162 cm So, the equation of line of best fit should be $y=162$ where $y$ is the girls height in cms which is almost independent of $x$.

$textbf{(a)}$ A very less percentage of variability in girls height can be predicted by their score in the Chapter 5 Test.

$textbf{(b)}$ The line of best best should be a horizontal line dividing the completely scattered points in equal halves so that the residual magnitudes and the number of points above and below to the line of best fit should be same.

As it is given that the average height of girls in her cass is 162 cm So, the equation of line of best fit should be $y=162$ where $y$ is the girls height in cms which is almost independent of $x$.

$textbf{(a)}$ The LSRL equation (calculated via excel) is
$$
y=1.3571x+7.7411
$$

where $y$ is the price of pizza in dollars and $x$ represents the number of topping on the pizza.

$y$-intercept in this context represents the price of a pizza if there is 0 toppings on it.

$textbf{(b)}$ Correlation coefficient $(r)$ for this data is $0.8616$, and hence the coefficient of determination $R^2$ for this data will be

$$
R^2=rtimes r=0.8616^2=0.7423
$$

$textbf{(c)}$ The slope of of the LSRL is positive and hence the association between price and the toppings on the pizza is positive which means as the toppings on the pizza increases the price also increases.

The value of $R^2$ is $0.7423$ which means approximately 74 percent of variation is price can be predicted by number of toppings on the pizza. Which signifies that the strength of the association is strong.

$textbf{(a)}$ The LSRL equation (calculated via excel) is $y=1.3571x+7.7411$

$textbf{(b)}$ Correlation coefficient $(r)$ for this data is $0.8616$, and hence the coefficient of determination $R^2$ for this data will be

$$
R^2=rtimes r=0.8616^2=0.7423
$$

$textbf{(c)}$ The slope of of the LSRL is positive and hence the association between price and the toppings on the pizza is positive. The value of $R^2$ is $0.7423$ which means approximately 74 percent of variation is price can be predicted by number of toppings on the pizza. Which signifies that the strength of the association is strong.

$textbf{(a)}$ Gulia’s father charges
$$
7.00 for a cheese pizza plus $1.50 for each additional topping.\
The points for the scatterplot can be calculated as shown below
begin{align*}
text{Pizza with 0 toppings }&rightarrow 7 \
text{Pizza with 1 topping }&rightarrow 7+1.5times 1=8.5\
text{Pizza with 2 toppings }&rightarrow 7+1.5times 2=10\
text{Pizza with 3 toppings }&rightarrow 7+1.5times 3=11.5\
text{Pizza with 4 toppings }&rightarrow 7+1.5times 4=13
end{align*}
Let
$$
x$represents the number of toppings on the pizza and$y$represent the price of pizza in $ then the scatterplot for this data can be drawn and LSRL equation can be calculated as shown on the plot below.\The LSRL equation from the plot is$y=1.5+7$\ The slope of LSRL is$1.5$which represent the change in the price of pizza per increase in the number of topping and$y$-intercept for this line is$7$ which represents the price of pizza if there is not any topping on the pizza.

$textbf{(b)}$ The value of both $r$ and $R^2$ is 1 as there is perfectly linear relationship between price and number of toppings on the pizza and 100 percent variation in price can be predicted by the number of topping on the pizza.

The variation in the previous question was 74 percent accurately predicted by the variation in the number of toppings because it was not a perfectly linear relationship and In problem 4-95 there may the other factors also which decides the price of a pizza.

$textbf{(a)}$ The LSRL equation from the plot is $y=1.5+7$
The slope of LSRL is $1.5$ which represent the change in the price of pizza per increase in the number of topping and $y$-intercept for this line is $7$ which represents the price of pizza if there is not any topping on the pizza.

$textbf{(b)}$ The value of both $r$ and $R^2$ is 1 as there is perfectly linear relationship between price and number of toppings on the pizza and 100 percent variation in price can be predicted by the number of topping on the pizza.

The variation in the previous question was 74 percent accurately predicted by the variation in the number of toppings because it was not a perfectly linear relationship and In problem 4-95 there may the other factors also which decides the price of a pizza.

Completely Describing Association
Description of Association in a scatterplot should always include description of the following.

$bullet$ Form $rightarrow$ Which tells about linearity or non-linearity which can be predicted seeing the shape of the graph as well as coefficient of correlation $(r)$

$bullet$ Direction $rightarrow$ Which tells about if the association is positive or negative and can be seen via graph and numerically detemined using the slope of LSRL.

$bullet$ Strength $rightarrow$ Which tells about if the association is strong, weak or not at all. This can be numerically determined using the value of $R^2$

Using the example in the problem 4-95 We can see that the equation of LSRL is
$$
y=1.3571x+7.7411
$$

where $y$ is the price of pizza in dollars and $x$ represents the number of topping on the pizza.

$bullet$ The slope of LSRL is positive and hence $textbf{Direction}$ in this case is positive.

$bullet$ The value of $r$ is $0.8616$ which is close to 1 which signifies that the $textbf{Form}$ in this case is almost linear.

$bullet$ The value of $R^2$ is $0.7423$ which signifies that the $textbf{Strength}$ in this case is strong as approximately 74 percent of variation is price can be predicted by number of toppings on the pizza.

textbf{(a)} The LSRL equation can be calculated by the procedure explained below.\ First we need to calculate the values for predefined variables $a$ and $b$ which is defined the in terms of $x$ and $y$ which is shown below.\$$a=dfrac{(sum y)(sum x^2)-(sum x)(sum xy)}{n(sum x^2)-(sum x)^2}$$\
$$b=dfrac{n(sum xy)-(sum x)(sum y)}{n(sum x^2)-(sum x)^2}$$
In the terms of $a$ and $b$ the LSRL equation is given by $$y=bx+a$$
The table of values is attached below.\\
begin{tabular}{ |p{1.5cm}|p{3cm}|p{3cm}|p{1cm}|p{1cm}|p{1cm}| }
hline
Serial & supplement $(x)$ & cold lasted $(y)$ &$xy$ & $x^2$ & $y^2$\
hline
1 & 0.5 & 4.5 &2.25 & 0.25 & 20.25\
hline
2& 2.5 & 1.6 &4 & 6.25 & 2.56\
hline
3 & 1& 3 &3 & 1 & 9\
hline
4 & 2 &1.8 &3.6 & 4 & 3.24\
hline
5 & 0.5 & 5 &2.5 & 0.25 & 25\
hline
6 & 1 & 4.2 &4.2 & 1 & 17.64\
hline
7 & 2 & 2.4 &4.8 & 4 & 5.76\
hline
8 & 1 & 3.6 &3.6 & 1 & 12.96\
hline
9 & 1.5 &3.3 &4.95 & 2.25 & 10.89\
hline
10 & 2.5 & 1.4 &3.5 & 6.25 & 1.96\
hline
$sum$ & 14.5 & 30.8 &36.4& 26.25 & 109.26\
hline
end{tabular}

Now, we need to calculate the values of $a$ and $b$ by using the values from the table above.

$$
begin{align*}
a&=dfrac{(30.8)(26.25)-(14.5)(36.4)}{10(26.25)-(14.5)^2}\
a&=dfrac{808.5-527.8}{262.5-210.25}\
a&=dfrac{280.7}{52.25}\
a&=5.372248\
a&=5.3722tag{On estimating }
end{align*}
$$

$$
begin{align*}
b&=dfrac{10(36.4)-(14.5)(30.8)}{10(26.25)-(14.5)^2}\
b&=dfrac{364-446.6}{26.5-210.25}\
b&=dfrac{-82.6}{52.25}\
b&=-1.5808612\
b&=-1.5809tag{On estimating }
end{align*}
$$

Therefore, After putting $a=5.3722$ and $b=-1.5809$, the final equation of LSRL becomes

$$
boxed{y=-1.5809x+5.3722}
$$

The scatterplot along with the electronically (via Excel) calculated value is attached below.

$textbf{(b)}$ Yes, this linear model is appropriate as the value of residuals are small and also we can numerically verify it by value of correlation coefficient. We can see that value of $rapprox -0.95$ which is very close to -1 which means that the model is strong linearly decreasing.

$textbf{(c)}$ The electronically (via excel) calculated value of correlation coefficient $(r)$ is $-0.95239$ and the value of $R^2$ is $0.9070515$

$textbf{(d)}$ There is strong, negative and linear association between the number of days cold lasted and the number of months taking supplements.

$bullet$ The form here is linear which can be verified by the value of $r$ which is very close to $-1$ and hence linear.

$bullet$ The direction is negative which can be verified by the slope of the LSRL equation which is $-1.5809$ and hence direction is negative

$bullet$ The strength is strong which can be verified by the value of $R^2$ which is $0.9070515$ which signifies that the 90 percent variation in number of days cold lasted can be predicted via number of days supplements taken. Which tells that the association is strong.

$bullet$ Data point is $(1,3)$ is an outlier as the residual for this is bigger than the trend.

$textbf{(a)}$ The equation of LSRL $y=-1.5809x+5.3722$

$textbf{(b)}$ Yes, this linear model is appropriate as the value of residuals are small and also we can numerically verify it by value of correlation coefficient. We can see that value of $rapprox -0.95$ which is very close to -1 which means that the model is strong linearly decreasing.

$textbf{(c)}$ The electronically (via excel) calculated value of correlation coefficient $(r)$ is $-0.95239$ and the value of $R^2$ is $0.9070515$

$textbf{(d)}$ There is strong, negative and linear association between the number of days cold lasted and the number of months taking supplements.

For individual explanation see inside.

$textbf{(a)}$
$$
begin{align*}
&dfrac{x^{150}}{x^{50}}tag{use $dfrac{a^{m}}{a^{n}}=a^{m-n}$ }\
&=x^{150-50}\
&=x^{100}
end{align*}
$$

Therefore, Gerardo’s rewritten expression $dfrac{x^{150}}{x^{50}}=x^3$ is incorrect.

$textbf{(b)}$
$$
begin{align*}
&y^{20}cdot y^{41}tag{use $a^mcdot a^n=a^{m+n}$}\
&=y^{20+41}\
&=y^{61}
end{align*}
$$

Therefore, Gerardo’s rewritten expression $y^{20}cdot y^{41}=y^{61}$ is correct.

$textbf{(c)}$
$$
begin{align*}
&(2m^2n^{15})^3tag{use $(a^m)^n=a^{mcdot n}$}\
&=2^{1cdot 3}m^{2cdot 3}n^{15cdot 3}\
&=2^{3}m^{6}n^{45}\
end{align*}
$$

Therefore, Gerardo’s rewritten expression $(2m^2n^{15})^3=2m^{6}n^{45}$ is incorrect.

$textbf{(a)}$ Incorrect

$textbf{(b)}$ Correct

$textbf{(c)}$ Incorrect

The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Therefore, the equation of a line of slope $-dfrac{4}{3}$ passing through a point $(12,-4)$ is given by

$$
begin{align*}
y-(-4)&=-dfrac{4}{3}cdot (x-12)\
y+4&=-dfrac{4}{3}cdot x -dfrac{4}{3}cdot (-12)\
y+4&=-dfrac{4}{3}cdot x+dfrac{48}{3}\
y+4&=-dfrac{4}{3}cdot x+16\
y&=-dfrac{4}{3}cdot x+16-4\
&boxed{y=-dfrac{4}{3}cdot x+12}\
end{align*}
$$

$$
y=-dfrac{4}{3}cdot x+12
$$

We can work backwards to ensure that the equation we will write has no solution.

Start with a false equation:

$$
1=0
$$

Using different properties of equality, we can rewrite this equation so that it has a variable $x$. One possible way is as follows:

Add $x$ to both sides:

$$
x+1=x
$$

Multiply both sides by 2:

$$
2x+2=2x
$$

Using the distributive property, we can rewrite the left side:

$$
color{#c34632}2(x+1)=2x
$$

Possible answer: $2(x+1)=2x$ ; Hint: We can work backwards to ensure that the equation we will write has no solution.

Domain of a function is a set of values of the inputs for which the function is defined. Therefore, for a given graph the value of variable $x$ for which the given function is defined and gives output in $y$, constitute its domain.

$textbf{(a)}$ D: all real numbers

The graph 1) takes all values of $x$ as input, therefore its domain is all real numbers.

$textbf{(b)}$ D:$x > -2$

The graph 3) is defined only for values of $x>-2$ as input, therefore its domain is all real numbers.

$textbf{(c)}$ D: $x leq 3$

The graph 2) corresponds for this domain for the similar reasons mentioned above.

$textbf{(a)}rightarrow$ Graph 1)

$textbf{(b)}rightarrow$ Graph 3)

$textbf{(c)}rightarrow$ Graph 2)