$text{textcolor{#4257b2}{textbf{c. }}}$

Joining your results with the rest of your class will increase your number of trials, and the experimental probability will gradually approach the theoretical probability of 93.75% (see Problem 6-3 on this page for details on how to reach this number).

$textbf{(b)}$ According to experiment carried above probability of a couple having a girl if they try
until they have a girl is:

$$
dfrac{textrm{Number of successful events }}{textrm{Total number of events}}=dfrac{24}{25}=0.96=textcolor{#4257b2}{96%}
$$

With a larger data set probability must be closer to theoretical probability i.e. $93.75%$

This is better estimate as coin can be unevenly weighted or have some other defects which may favour a side more than other whereas in this method numbers are generated randomly without any bias.

For every child born, there is a $dfrac{1}{2}$ probabilitiy that the child is a girl.

The probability of 4 boys is then: $left(dfrac{1}{2}right)^4=dfrac{1}{16}$.

The probability of first 3 boys and then a girl: $left(dfrac{1}{2}right)^4=dfrac{1}{16}$.

The probability of first 2 boys and then a girl: $left(dfrac{1}{2}right)^3=dfrac{1}{8}$.

The probability of first 1 boy and then a girl; $left(dfrac{1}{2}right)^2=dfrac{1}{4}$.

The probability of the first child to be a girl is $left(dfrac{1}{2}right)$.

Thus the probability of having a girl is then:

$$
dfrac{1}{16}+dfrac{1}{8}+dfrac{1}{4}+dfrac{1}{2}=dfrac{15}{16}=93.75%
$$

a. 25% of the homes do not have large backyards and 20% do not have garages. The probability of both is then the product:

$$
25%cdot 20% = 0.25cdot 0.2=0.05=5%
$$

b. Sine the probability of having a garage is independent of having a large back yard, the conditional probability of having a large backyard is equal to the probability of having a large backyard 75%.

$log(0.3)textless1$ since the solution $x=log(0.3)$ is $10^{x}=0.3$. If we raise raise 10 to a power x and generate a number less than unity, then x must be negative

It is given that,
$$begin{aligned}
text{Radius of the cork}&= 2 text{cm}\
text{Height of the cork}&= 5 text{cm}\
text{Weight of the cork}&= 2.5 text{grams}\
end{aligned}$$

Determining Volume of the cork:
$$begin{aligned}
text{Volume}&= pi r^2 h\
end{aligned}$$
Substituting the values:
$$begin{aligned}
&= dfrac{22}{7}cdot (2)^2 cdot 5 \
&= 3.14cdot 4 cdot 5 \
&boxed{text{Volume}= 62.83 text{cm}^3} \
end{aligned}$$

Determining density of the cork:
$$begin{aligned}
text{density}&= dfrac{text{Mass}}{text{Volume}}\
end{aligned}$$
Substituting the values:
$$begin{aligned}
&= dfrac{2.5}{62.83}\
&boxed{text{density}= 0.0397 dfrac{text{g}}{text{cm}^3}}
end{aligned}$$

Volume : 62.83 $text{cm}^3$

Density : $0.0397 dfrac{text{g}}{text{cm}^3}$

Given,
$$begin{aligned}
f(x)&= 2x^2 – 3x + 1rightarrow(1)\
g(x)&= 4x-2rightarrow(2)\
end{aligned}$$

Equating equation (1) and (2):
$$begin{aligned}
2x^2 – 3x + 1&= 4x-2\
2x^2 – 3x -4x+1+2&=0\
2x^2 – 7x+3&=0rightarrow(3)\
end{aligned}$$

Factorizing equation (3):
$$begin{aligned}
2x^2- 6x – x+3&=0\
2x(x-3)-1(x-3)&=0\
(2x-1) (x-3)&=0rightarrow(4)\
end{aligned}$$

Solving further:
$$begin{aligned}
2x-1&=0\
2x&=1\
&boxed{x=dfrac{1}{2}}\
end{aligned}$$
Also,
$$begin{aligned}
x-3&=0\
&boxed{x=3}\
end{aligned}$$

$x= dfrac{1}{2} ; 3$

$textbf{(a)}$ If number of Black outcomes and Red outcomes are not equal then the wheel must be rigged as theoretical probability of getting either Red or Black is $50%$

$textbf{(b)}$ Below are results of 200 experiments:

$$
B, R, B, R, B, B, B, B, B, R, R, R, B, B, B, B, R, R, B, R,
$$

$$
B, B, B, R, R, R, B, B, B, R, B, R, R, R, B, R, R, R, B, R,
$$

$$
R, B, R, B, R, R, R, R, R, R, R, R, B, R, R, R, R, R, R, B,
$$

$$
B, R, B, B, B, B, R, B, R, B, R, B, R, B, B, R, B, R, B, B,
$$

$$
B, R, R, B, B, R, B, B, B, B, R, B, R, R, B, B, B, R, R, R,
$$

$$
B, R, B, B, R, R, R, B, B, B, B, R, B, B, R, B, R, R, R, B,
$$

$$
R, R, B,R, B, B, R, R, R, R, B, R, B, B, B, B, R, R, R, B,
$$

$$
R, R, B, B, B, B, B, R, B, B, B, B, R, B, R, R, B, B, R, B,
$$

$$
B, B, R, B, R, R, B, R, R, R, B, B, R, B, B, B, B, B, B, B,
$$

$$
R, B, B, B, B, R, B, B, B, R, B, R, B, R, R, B, R, R, B, B
$$

Total number of Red outcomes is: 93

Total number of Black outcomes is: 107

$textbf{(c)}$ It can be observed that there are 2 streaks of five or more reds and 3 streaks of five or more blacks.

$textbf{(d)}$ Wheel might be rigged if more number of streaks or longer streaks for a color are observed compare to other color

$textbf{(e)}$ Yes, it will be to advantage of owner to leave it that way as it changes the probability of getting Red or Black and can change payouts for the offer with lower odds of winning.

$textbf{(a)}$ Below are results obtained from 200 experiments:

* $bullet$ Number of times Figure 1 is obtained: 31
* $bullet$ Number of times Figure 2 is obtained: 30
* $bullet$ Number of times Figure 3 is obtained: 29
* $bullet$ Number of times Figure 4 is obtained: 38
* $bullet$ Number of times Figure 5 is obtained: 28
* $bullet$ Number of times Figure 6 is obtained: 44

It can be observed Figure 5 have least probability i.e $0.14$

Expected value of getting Figure 5 is $dfrac{1}{textrm{Probability}}=$ 7.14

Hence, Katelyn should buy $text{textcolor{#4257b2}{8 meals}}$ to get all 6 action figures

$textbf{(b)}$ Below are results obtained after 10 more simulations:

* $bullet$ Number of times Figure 1 is obtained: 346
* $bullet$ Number of times Figure 2 is obtained: 374
* $bullet$ Number of times Figure 3 is obtained: 346
* $bullet$ Number of times Figure 4 is obtained: 359
* $bullet$ Number of times Figure 5 is obtained: 397
* $bullet$ Number of times Figure 6 is obtained: 378

It can be observed Figure 1 and Figure 3 have least probability i.e $0.1572$

Expected value of getting Figure 1 and Figure 3 is $dfrac{1}{textrm{Probability}}=$ 6.3613

Hence, Katelyn should buy $text{textcolor{#4257b2}{7 meals}}$ to get all 6 action figures

$textbf{(c)}$ After combining with team, the answer must be more closer to 6

$textbf{(d)}$ The least number of meals to buy to get all six figures is 6. Whereas maximum number can be any number greater than equal to 6 as the it is not certain but only a possibility of getting all six figures

It is possible to buy 50 or 100 meals to get all six figures as there is no certain value for maximum number of meals to be bought

We note in the table that Myriah has a probability of $dfrac{15}{36}$ to do the dishes (two dice give sum of less than or equal to 6) and $dfrac{21}{36}$ to not have to do the dishes.

The number of days in the week that she can expect to wash the dishes is then:
$$
7cdot dfrac{15}{36}=2.916667approx 3
$$

Thus about 3 days.

Express the argument of the log function as a power of the base 25 in (a) $x=log_{25}25^{dfrac{1}{2}}=dfrac{1}{2}log_{25}25=dfrac{1}{2}$

(b) $x$ does not have a unique solution but can take on any value in the interval $left( 0,1right)cupleft( 1,inftyright)$ Note $xne1$ due to division by zero in the change of base formula

(c) $x=10^{23}$

$$
3^x+5=x^2-5
$$

$$
xapprox -3.167
$$

$$
xapprox -3.167
$$

$$
|x|<3
$$

$$
-3<x<3
$$

$(-3,3)$

$$
|2x+1|<3
$$

$-3<2x+1<3$

$-3-1<2x+1-1<3-1$

$-4<2x<2$

$$
-2<x<1
$$

$$
|2x+1|geq 3
$$

$2x+1leq -3$ or $2x+1geq 3$

$2x+1-1leq -3-1$ or $2x+1-1geq 3-1$

$2xleq -4$ or $2xgeq 2$

$xleq -2$ or $xgeq 1$

$(-infty, -2]cup[1,infty)$

a) $(-3,3)$

b) $(-2,1)$

c) $(-infty, -2]cup[1,infty)$

$$
f(x)=dfrac{1}{2}(x+1)^3
$$

We sketch the graph of $f(x)$ starting with the parent function $y=x^3$ which we shift 1 unit to the left and the vertically shrink by a factor of $dfrac{1}{2}$:

We sketch the graph of the inverse $f^{-1}(x)$ by reflecting $f(x)$ across the line $y=x$:

$y=dfrac{1}{2}(x+1)^3$

$x=dfrac{1}{2}(y+1)^3$

$2x=(y+1)^3$

$y+1=sqrt[3]{2x}$

$y=sqrt[3]{2x}-1$

$$
f^{-1}(x)=sqrt[3]{2x}-1
$$

$$
f^{-1}(x)=sqrt[3]{2x}-1
$$

$textbf{(a)}$ Proportion of blue candies can be given as:

$$
dfrac{textrm{Number of blue candies}}{textrm{Total number of candies}}
$$

Simulating such a scenario expecting to get 50 candies per package we have below result:

Brown: 8

Yellow:10

Red:5

Orange:10

Green:7

Blue.10

Hence, for above simulation proportion of blue candies is $dfrac{10}{50}=textcolor{#4257b2}{0.2}$

textbf{(b)} Expecting a strength of around 25 students in class and repeating the experiment 24 times more we have below results:
$$0.20,
0.20,
0.16,
0.08,
0.14,
0.26,
0.12,$$
$$0.18,
0.08,
0.12,
0.2,
0.16,
0.20,
0.10,$$
$$
0.26,
0.22,
0.28,
0.28,
0.14,
0.18,$$
$$
0.08,
0.16,
0.22,
0.16,
0.12
$$
below table:\\
begin{tabular}{|c|c|}
hline
textbf{Class} & textbf{Frequency} \ hline
$0.08-0.10$ & 3 \ hline
$0.10-0.12$ & 1 \ hline
$0.12-0.14$ & 3 \ hline
$0.14-0.16$ & 2 \ hline
$0.16-0.18$ & 4 \ hline
$0.18-0.20$ & 2 \ hline
$0.20-0.22$ & 4 \ hline
$0.22-0.24$ & 2 \ hline
$0.24-0.26$ & 0 \ hline
$0.26-0.28$ & 2 \ hline
$0.28-0.30$ & 2 \ hline
end{tabular}\\
Below is histogram for above frequency table:

$textbf{(c)}$ Evaluating mean$(overline{x})$ proportion of class:

$$
begin{align*}
overline{x}&=dfrac{4.3}{25}\
&=textcolor{#4257b2}{0.172}
end{align*}
$$

Mean proportion gives estimate of expected value of proportion received in a random packet

$textbf{(d)}$ We expect the proportion of blue candies in the sample to be between $text{textcolor{#4257b2}{underline{0.08}}}$ and $text{textcolor{#4257b2}{underline{0.3}}}$.

$textbf{(e)}$ Proportion of blue candies in whole population is $color{#4257b2}{0.172 pm 0.11}$

Margin of error is 0.11

(c) Mean is expected proportion received in a random packet (e) $0.172 pm 0.11$

textbf{(a)} Below are results obtained for 25 simulations:\\
begin{tabular}{|c|c|c|}
hline
textbf{Number of simulation} & textbf{Results} & textbf{Number of matches} \ hline
1 & $2,2,1,1,1,2,1$ & 7 \ hline
2 & $2,2,1,2,2$ & 5 \ hline
3 & $1,2,1,1,2,1$ & 6 \ hline
4 & $2,2,1,1,1,2,2$ & 7 \ hline
5 & $2,1,2,2,1,2$ & 6 \ hline
6 & $2,2,2,1,1,1,1$ & 7 \ hline
7 & $2,2,2,1,1,1,2$ & 7 \ hline
8 & $1,1,1,2,1$ & 5 \ hline
9 & $2,1,1,2,1,2,2$ & 7 \ hline
10 & $1,2,2,2,1,2$ & 6 \ hline
11 & $2,2,2,1,2$ & 5 \ hline
12 & $1,2,1,1,1$ & 5 \ hline
13 & $2,1,1,2,1,1$ & 6 \ hline
14 & $2,2,2,2$ & 4 \ hline
15 & $1,1,1,1$ & 4 \ hline
16 & $1,1,2,2,1,2,2$ & 7 \ hline
17 & $2,2,2,1,1,1,2$ & 7 \ hline
18 & $2,1,1,1,1$ & 5 \ hline
19 & $1,1,2,2,2,1,2$ & 7 \ hline
20 & $1,2,1,2,2,2$ & 6 \ hline
21 & $1,1,1,2,1$ & 5 \ hline
22 & $2,2,2,2$ & 4 \ hline
23 & $1,2,1,2,1,1$ & 6 \ hline
24 & $1,1,2,1,1$ & 5 \ hline
25 & $2,2,2,2$ & 4 \ hline
end{tabular}

$textbf{(b)}$ Average number of games a World Series lasts is:

$$
dfrac{143}{25}=textcolor{#4257b2}{5.72 textrm{ games}}
$$

(b) $5.72 textrm{ games}$

textbf{(a)} Let textcolor{blue}{$1$} denote textcolor{blue}{successful throw} and textcolor{red}{$0$} denote textcolor{red}{unsuccessful throw}.\\ Below are results after 25 simulations of 20 free throws:\\
begin{tabular}{|c|c|c|}
hline
textbf{Number of simulation} & textbf{Results} & textbf{Longest streak} \ hline
1 & $1,1,0,0,0,0,1,1,1,0,0,0,0,0,1,1,1,0,0,0$ & 3 \ hline
2 & $1,1,0,1,1,1,0,1,1,1,1,0,0,0,0,1,0,1,1,1$ & 4 \ hline
3 & $0,0,1,1,0,1,1,1,0,0,0,1,1,0,0,1,1,1,0,1$ & 3 \ hline
4 & $1,1,0,0,1,1,0,1,0,0,0,0,0,0,0,1,1,1,0,0$ & 3 \ hline
5 & $0,0,1,0,0,0,0,0,1,0,0,0,1,1,1,1,0,1,0,1$ & 4 \ hline
6 & $1,1,0,1,0,1,0,1,0,0,1,1,0,0,1,0,1,1,0,1$ & 2 \ hline
7 & $1,0,1,1,1,1,0,0,1,0,0,0,1,1,0,0,1,0,0,1$ & 4 \ hline
8 & $1,1,0,1,0,0,1,1,1,0,1,0,1,1,1,0,0,1,0,1$ & 3 \ hline
9 & $0,0,1,0,1,1,1,1,1,0,1,1,1,0,0,1,0,0,0,1$ & 5 \ hline
10 & $1,0,1,0,0,0,1,1,0,1,1,0,0,1,1,1,0,1,0,0$ & 3 \ hline
11 & $1,0,1,0,1,0,1,0,1,0,1,1,1,0,0,1,1,0,0,0$ & 3 \ hline
12 & $1,0,0,0,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1,1$ & 3 \ hline
13 & $0,1,0,1,0,0,1,1,1,1,0,1,0,0,1,1,0,1,1,1$ & 4 \ hline
14 & $0,0,0,0,1,0,1,0,0,1,0,1,0,0,0,0,1,0,1,1$ & 2 \ hline
15 & $1,1,0,0,0,1,1,0,1,0,1,1,0,1,1,0,1,1,0,0$ & 2 \ hline
16 & $0,0,0,1,0,0,1,1,1,1,1,0,0,0,0,1,0,1,1,0$ & 5 \ hline
17 & $1,1,0,1,0,1,0,0,0,0,0,1,1,1,0,1,0,0,0,0$ & 3 \ hline
18 & $0,1,1,0,0,0,1,1,0,0,1,1,0,0,1,0,1,0,1,0$ & 3 \ hline
19 & $1,0,1,1,1,0,0,1,1,1,0,0,0,0,1,0,1,1,0,0$ & 3 \ hline
20 & $1,1,0,1,0,0,1,1,0,1,1,1,0,0,0,0,1,1,1,1$ & 4 \ hline
21 & $1,1,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,0,1,1$ & 2 \ hline
22 & $0,1,0,1,0,1,1,1,1,1,0,1,1,0,0,0,1,1,0,0$ & 5 \ hline
23 & $0,1,1,0,1,0,1,1,0,0,0,1,0,0,0,1,1,1,0,1$ & 3 \ hline
24 & $0,0,1,1,1,0,0,1,0,0,1,0,1,1,0,1,0,0,0,1$ & 3 \ hline
25 & $0,1,1,1,1,0,0,1,0,0,0,1,1,1,1,1,1,0,0,0$ & 6 \ hline
end{tabular}

$textbf{(b)}$ Average length of longest streak is:

$$
dfrac{85}{25}=textcolor{#4257b2}{3.4}
$$

The streak must be $text{textcolor{#4257b2}{at least 4}}$ successful throws long before it is considered unusual.

There are 38 possible fields, where we have 18 red and black fields, and 2 green fields. Probabilty of a ball landing on red is the as the probability of landing on black which is

$$
frac{18}{38},
$$

while the probability of a ball landing on green is

$$
frac{2}{38}.
$$

$textbf{a.}$

Make an area model as below:

$textbf{b.}$

Probability of the ball landing on red twice in a row is represented by an area of a RR rectangle which is

$$
frac{18}{38}cdot frac{18}{38}=frac{9}{19} cdot frac{9}{19} = frac{81}{361} approx boldsymbol{22.44%}.
$$

$textbf{c.}$

Probability of the ball landing on red on the second spin is represented by a sum of areas of RR, BR, and GR triangles, thus:

$$
frac{18}{38}cdot frac{18}{38}+frac{18}{38}cdot frac{18}{38}+frac{2}{38}cdot frac{18}{38}=frac{81}{361}+frac{81}{361}+frac{9}{361}=frac{171}{361}approx boldsymbol{47.37%}.
$$

$textbf{d.}$

Since it is given that the first spin was red, we consider only the first column. The required probability is the ratio of areas of RR rectangle and the sum of areas of RR, RB, RG rectangles.

$$
frac{frac{81}{361}}{frac{81}{361}+frac{81}{361}+frac{9}{361}} approx boldsymbol{47.37%}.
$$

$textbf{e.}$

They are the same because the probability of the ball landing on red is a single spin is $frac{18}{38}$, or $approx boldsymbol{47.37%}.$

$textbf{a.}$ Make the area model as was done throughout the chapter.

$textbf{b.}$ $approx 22.44%$

$textbf{c.}$ $approx 47.37%$

$textbf{d.}$ $approx 47.37%$

$textbf{e.}$ Probabilites are the same.

The function $f(x)=log_{5}(x-2)$ shown is the parent function $f(x)=log_{5}(x)$ shifted horizontally such that the x-intercept and vertical asymptote is 2 units further to the right

Replace $y$ with $dfrac{1}{2}$ in the second equation:

$$
dfrac{1}{2}=dfrac{16}{x^2-4}
$$

Use cross multiplication:

$$
(1)(x^2-4)=(16)(2)
$$

Simplify:

$$
x^2-4=32
$$

Add 4 to both sides of the equation:

$$
x^2=36
$$

Take the square root of both sides of the equation:

$$
x=pm 6
$$

$$
x=pm 6
$$

You can use your graphing calculator to simulate 25 students’ opinion about the dance. We are going to use MS Excel tool to show the sheet clearly. You can try any spreadsheet software yourself. Type the command $texttt{=RANDBETWEEN(1;100)}$ into the function bar and copy-paste the command to all required cells. A random integer from 1 to 100 will be generated in each cell. Below is the my table for a common test of the group, and also 5 tests of my own and 5 tests of my partner.

Also, people who support the dance are highlighted in $color{#4257b2}text{blue}$, and those who do not are colored in $color{#c34632}text{red}$.

In the last, separated row, there are proportions of students who support the dance. With this, $textbf{a.}$ and $textbf{b.}$ are solved.

$textbf{c.}$

The mean of all proportions combined is

$$
frac{0.52+0.44+0.76+0.72+0.72+0.72+0.6+0.56+0.72+0.56+0.68}{11}approx boldsymbol{0.64},
$$

which is very close to $0.6$, which is we would expect. This is because $0.6$ is the expected proportion.

$textbf{d.}$

We have to find the middle $90%$ of all means, which means that the lower bound should be $0.52$, and the upper bound $0.72$.

$textbf{e.}$

The margin of error is half of range between the bounds, so the margin of error is

$$
frac{0.72-0.52}{2}=boldsymbol{0.1}.
$$

The required proportion is

$$
boldsymbol{0.64pm 0.1}.
$$

Conduct your own tests. They should produce similar results as given above. The proportion should be around $0.6pm 0.1$.

Refer to the test done in the previous problem.

With this, since the proportion obtained was $0.64pm 0.1$, even with the margin of error, the principal should have the convincing evidence.

$textbf{a.}$

Larger sample means more credible results.

$textbf{b.}$

If the results on the sample of 25 students were credible, and they should be since the mean of proportions was close to the expected value, then the margin of error should be similar. On a larger sample, it could be even smaller because the mean could be closer to the real proportion.

$textbf{c.}$

Let’s set up 5 simulations for 100 students. The results will be written as in the previous problem. People who support the dance are highlighted in $color{#4257b2}text{blue}$, and those who do not are colored in $color{#c34632}text{red}$.

$textbf{d.}$

The mean of all proportions combined is

$$
frac{0.66+0.55+0.56+0.56+0.65}{5}approx boldsymbol{0.596},
$$

which is extremely close to $0.6$, which is what we would expect. This is because $0.6$ is the expected proportion.

Comparing with the class, it turns out that the lower bound is $0.568$, and the upper bound $0.64$. In that case, the margin of error is

$$
frac{0.64-0.568}{2}=0.036
$$

$textbf{e.}$

If nothing else, the principal should be more convinced. The results gather even more around $0.6$ which is expected value and that means that by increasing the sample, results got more credible. The principal may argue that the sample was not representative enough, but surely, their results should not be discarded.

$textbf{a.}$

To find the mean proportion, we have to add all proportions and divide by 100 (a number of samples). Luckily, we are given the sum, 7.83, so the mean proportion is

$$
frac{7.83}{100}=boldsymbol{7.83%}.
$$

$textbf{b.}$

Out of 100 samples, the lowest 5 represent the lower $5%$, and the greatest 5 the upper $5%$. We have to dismiss these values, and they are 0.02, 0.04, 0.05, 0.05, 0.05, so the lower bound will be $boldsymbol{0.05}$. We also dismiss 0.11, 0.11, 0.12, 0.12, 0.13, so the upper bound is $boldsymbol {0.11}$.

$textbf{c.}$

The proportion of defective flashlights in the whole population is $boldsymbol{7.83%}$ with the margin of error $frac{0.11-0.05}{2}=0.03=boldsymbol{3%}$.

$textbf{a.}$ $7.83%$

$textbf{b.}$ $0.05$ and $0.11$

$textbf{c.}$ $7.83%pm 3%$

$textbf{a.}$

You could use your online studying materials to simulate dice rolling, or you could use your graphing calculators or computer software. We are going to use MS Excel tool. You can try any spreadsheet software yourself. Type the command $texttt{=RANDBETWEEN(1;6)}$ into the function bar and copy-paste the command to all required cells. A random integer from 1 to 6 will be generated in each cell, representing the dice roll. I will show my results below.

Myriah doing the dishes will be represnted with $color{#c34632}text{red}$ highlighting, and parents doing the dished with $color{#4257b2}text{blue}$ highlighting.

$textbf{a.}$ Use technology to run the simulation. You can add a column for a sum to make the relevant result easier to see.

$textbf{b.}$ This simulation have shown that she will have to do dishes 3 or more times in a row in 2 instances.

$a$.
$$begin{aligned}
y&= log_{12}(x)rightarrow(1)\
end{aligned}$$

Re-writing equation(1) in exponential form:
$$begin{aligned}
x&= 12^{y}
end{aligned}$$

$b$.
$$begin{aligned}
x&= log_{y}(17)rightarrow(2)\
end{aligned}$$
Re-writing equation(2) in exponential form:
$$begin{aligned}
y^x&= 17\
end{aligned}$$

$c$.
$$begin{aligned}
y&= 1.75^{2x}rightarrow(3)\
end{aligned}$$

Re-writing equation(3) in exponential form:
$$begin{aligned}
log_{1.75}(y)&= log_{1.75}(1.75)^{2x}\
log_{1.75}(y)&=2x log_{1.75}(1.75)\
log_{1.75}(y)&=2x\
end{aligned}$$

$c$.
$$begin{aligned}
3y&= x^7rightarrow(4)\
end{aligned}$$
Re-writing equation(4) in exponential form:
$$begin{aligned}
log_{x}(3y)&= log_{x}(x^7)\
log_{x}(3y)&=7\
end{aligned}$$

(a) using base 2 logarithm; $7=log_{2}128$. (b) It is true because $7=log_{2}128=k$ where k is a positive constant, we understand the logarithm of k is a constant.

$a$.
Equation to be solved by Jenna,
$$begin{aligned}
dfrac{3}{x-2} + dfrac{8}{10}&= dfrac{x}{x-2}rightarrow(1)\
end{aligned}$$

Multiply both sides by 10:
$$begin{aligned}
dfrac{3}{x-2} cdot 10 + dfrac{8}{10}cdot 10&= dfrac{x}{x-2}cdot10rightarrow(1)\\
&boxed{dfrac{30}{x-2} + 8=dfrac{10x}{x-2}rightarrow(2)}\\
end{aligned}$$
Hence, equation (2) is the new equation.

$b$.
Eliminating $dfrac{30}{x-2}$:

Subtracting $dfrac{30}{x-2}$ both sides,
$$begin{aligned}
dfrac{30}{x-2} – dfrac{30}{x-2} +8 – dfrac{30}{x-2} &= dfrac{10x}{x-2}-dfrac{30}{x-2}\\
8&= dfrac{10x-30}{x-2}\\
8(x-2)&= 10x-30\
8x-16&=10x-30\
8x-10x&= -30+16\
-2x&= -14\\
x&=dfrac{14}{2}\\
&boxed{x=7}
end{aligned}$$

Checking the answer by putting $x = 7$ at the left-hand side and right-hand side respectively.
Left-hand side:
$$begin{aligned}
&=dfrac{3}{7-2}+dfrac{8}{10}\\
&= dfrac{3}{5}+dfrac{4}{5}\\
&= dfrac{7}{5}rightarrow(a)\\
end{aligned}$$
Right-hand side:
$$begin{aligned}
&=dfrac{7}{7-2}\\
&=dfrac{7}{5}rightarrow(b)\
end{aligned}$$

From $(a)$ and $(b)$, it is clear $x = 7$ is the right answer.

$$x = 7$$

We created a simulation using MS Excel software. You can try any spreadsheet software yourself. Type the command $texttt{=RANDBETWEEN(1;100)}$ into the function bar and copy-paste the command to all required cells. A random integer from 1 to 100 will be generated in each cell.

Surviving tadpoles will be highlighted $color{#4257b2}text{blue}$ and dying $color{#c34632}text{red}$. As calculated, the proportion of surviving tadpoles is $boldsymbol 68%$.

$textbf{b.}$

The results of simulation are below. It is done using the same software, with the same conditions.

Surviving tadpoles will be highlighted $color{#4257b2}text{blue}$ and dying $color{#c34632}text{red}$. As calculated, the proportion of surviving tadpoles is $boldsymbol 56%$.

$textbf{c.}$

The difference between these proportions is $68%-56%=boldsymbol{12%}$.

$textbf{d.}$

A positive difference means that the proportion of surviving tadpoles who were mosquito fed is greater than the proportion of surviving non mosquito fed tadpoles.

A negative difference means the converse.

$textbf{e.}$

Out of 100 samples, the lowest 5 represent the lower $5%$, and the greatest 5 the upper $5%$. We have to dismiss these values, and they are $-0.07$, $-0.05$, $-0.04$, $-0.04$, $-0.02$, so the lower bound will be $boldsymbol{-0.01}$. We also dismiss 0.30, 0.30, 0.30, 0.33, 0.36, so the upper bound is $boldsymbol {0.29}$.

The margin of error is $frac{0.29-(-0.01)}{2}=boldsymbol{0.15}$.

$textbf{f.}$

Since the means of proportion was $0.14$, we can predict that the difference between surviving tadpoles who were mosquito fed, and those who were not is $boldsymbol{14% pm 15%}$.

$textbf{g.}$

The difference of zero means that the same proportion of tadpoles survived, no matter if they were mosquito fed or not.

$textbf{h.}$

While it appears that there is a difference, we cannot be entirely convinced because the margin of error is greater than the mean of differences.

$textbf{a.}$

Out of 100 samples, the lowest 5 represent the lower $5%$, and the greatest 5 the upper $5%$. We have to dismiss these values, and they are $0.51$, $0.52$, $0.54$, $0.54$, $0.56$, so the lower bound will be $boldsymbol{0.56}$. We also dismiss 0.74, 0.75, 0.76, 0.76, 0.77, so the upper bound is $boldsymbol {0.74}$.

$textbf{b.}$

The margin of error is

$$
frac{0.74-0.56}{2}=boldsymbol{0.09}.
$$

The simulation mean is $65%$, so the predicted proportion is $boldsymbol{65%pm 9%}$.

$textbf{a.}$ Lower bound is $56%$, and upper bound is $74%$.

$textbf{b.}$ The predicted proportion is $65%pm 9%$.

We can be confident that the number of passengers who lost their luggage is in the range between $1.3%$ and $4.7%$.

Since $1.3%$ of 1000 is 13, and $4.7%$ of 1000 is 47, we can predict that if they surveyed 1000 passengers, the number of those who lost their luggage will be between 13 and 47.

a-

$tan 44=dfrac {x}{28}$

$0.9657=dfrac {x}{28}$

$x=0.9657 times 28$

$$
x=27.04 mathrm { ~feet}
$$

b-

$cos 32=dfrac {150}{x}$

$0.848=dfrac {150}{x}$

$x=dfrac {150}{0.848}$

$$
x=176.88 mathrm { ~cm}
$$

c-

$sin 38=dfrac {x}{47}$

$0.6157=dfrac {x}{47}$

$x=0.6157 times 47$

$$
x=28.94 mathrm { ~meters}
$$

a-          $x=27.04 mathrm { ~feet}$

b-          $x=176.88 mathrm { ~cm}$

c-          $x=28.94 mathrm { ~meters}$

According to the study, the DST will decrease the energy using from $-1.3%$ (that is actually increased energy usage) to $1.7%$. If they save 9 million dollars for $1%$ of decrease, they will save up to $9cdot 1.7=boldsymbol{15.3}$ $textbf{million dollars}$ of $1.7%$ decrease. Also, since $1.3cdot 9=11.7$, they could end up spending additional $boldsymbol{11.7}$ $textbf{million dollars}$.

$textbf{a.}$

There are multiple ways how to do this. One way would be flipping a coin 3 times, because there are 8 different outcomes. We could map each possible outcome to a different student.

$textbf{b.}$

The answer is 3, and the method is described in $textbf{a.}$

$textbf{c.}$

Yes. We can apply the method described. If the outcome of tossing a coin 3 times correspond to the absent student, we can just repeat the process.

$textbf{a.}$ Flip a coin 3 times. There are 8 possible outcomes which we map to each student.

$textbf{b.}$ 3

$textbf{c.}$ Apply the method described above. If the outcome correspond to the absent student, repeat the process.

Solve by taking logarithms to base ten; (a) $x=dfrac{log(16)}{log(10)}=1.2$ ; (b) $x=dfrac{log(41)}{log(10)}=1.61$ (c) $x=dfrac{log(729)}{log(3)}=6$ ; (d) $x=dfrac{log(101)}{log(10)}=2.004$

(a) Rearranging and factoring by grouping; $x^{2}-3x+x-3<0rightarrow x(x+1)-3(x+1)<0rightarrow(x-3)(x+1)<0$. If $-1<x<3$ then $(x-3)(x+1)<0$

(b) Rearranging and factoring by grouping; $x^{2}-2x-x+2geq0rightarrow x(x-1)-2(x-1)rightarrow(x-2)(x-1)geq0$ and is satisfied for $xleq1$ and $xgeq2$

$-1<x<3$ and $xgeq2$; $xleq1$

$textbf{a.}$

You could use your online studying materials to simulate dice rolling, or you could use your graphing calculators or computer software. We are going to use MS Excel tool. You can try any spreadsheet software yourself. Type the command $texttt{=RANDBETWEEN(1;6)}$ into the function bar and copy-paste the command to all required cells. A random integer from 1 to 6 will be generated in each cell, representing the dice roll. I will show my results below.

The mean turned out to be $2.9$.

$textbf{b.}$

Since there would be 2 people working in a group, each one of us made a simulation and recorded the result. The mean can also be seen for each sample, below the list of outcomes.

$textbf{c.}$

We will pretend that the class has 6 people. That would be 30 testing since everyone produces five simulations. Record the data in the table and draw a histogram of means (otherwise it would be too tedious to make a histogram of all data). The mean is found to be $3.58$.

$textbf{d.}$

Discard bottom $5%$ and top $5%$ of means. The lower bound is $2.5$ and the upper bound is $4.3$

a. It is given that that the population of movie goers is actually split between those preferring romantic comedies and those preferring action movies.

Thus, it can be inferred that the probability that a person prefers an action movie is $dfrac{1}{2}$. Similarly, the probability that a person prefers a romantic movie is also $dfrac{1}{2}$.

The question asks what are the chances that a group of 4 will prefer one genre over the other unanimously.

Accordingly, each of the 4 persons will have a 50% chance that she prefers action or romantic movies.

Thus,

$$begin{aligned}
text{P(group of 4 will prefer one genre over the other)} &= dfrac{1}{2} times dfrac{1}{2} times dfrac{1}{2} times dfrac{1}{2} \\
&=dfrac{1}{16}
end{aligned}$$
Therefore, there is a $6.25 % left(dfrac{1}{16} times 100right)$ chance that a group of 4 persons will prefer one genre over the other unanimously.

The conclusion arrived is plausible as one person’s preference is independent of the other person’s preference. Also, the plausibility is based on the assumption that there was no preconceived notion attached to any one genre.

We have to find the 5% upper and lower bound. for this 5% of the data needs to be excluded from above and from below. Total number of regions is 40, therefore, 5% of 40 i.e. 2 datasets have to be excluded from above and from below.

Accordingly, $24551, 24574$ are to be excluded from the first column. And we get the lower bound i.e. $24600$.
Similarly, $25249, 25041$ are to be excluded from the bottom of the last column. And we get the upper bound i.e. $25033$

We are given the total number of copies (total of all the 40 regions). The given total is 992,440.

Accordingly we can find the mean ($bar{x}$)

$$bar{x} = dfrac{992440}{40} = 24,811$$

We have to find the margin of error.

$$text{Margin of Error} = dfrac{text{Upper Bound – Lower Bound}}{2}$$

Calculating the margin of error,

$$begin{aligned}
text{Margin of Error}&=dfrac{25033-24600}{2} \\
&=dfrac{433}{2} \\
&=216.5
end{aligned}$$

The expected number of copies that can be made from a copier before it requires maintenance is

$$24,811 pm 216.5$$

The range is thus from $24594.5$ to $25027.5$. Since the company has claimed that their machines can copy at least 25000 copies the researchers may be successful in filing their claim.

$textbf{a.}$

The difference in proportions is

$$
84% – 72% = 12% = boldsymbol{0.12}
$$

$textbf{b.}$

The difference of zero is not a plausible result, because 0 is not in the confidence interval. A difference of zero means that there is no difference whether the detergent is used, or not.

$textbf{c.}$

Since the proportion of difference is positive, even with margin of error included, we can say that we are confident that there is a difference between washing the dishes with detergent and doing this without it.

$textbf{a.}$ 0.12

$textbf{b.}$ No. Zero means that there is difference whether detergent is used, or not.

$textbf{c.}$ Yes.

Given,
$$log(x) = log (5)rightarrow(1)$$

In order to check whether the equation (1) and $x = 5$ are equivalent or not, we will simplify the equation (1) by taking antilog both sides:
$$begin{aligned}
log(x)&= log(5)\
&boxed{x=5}
end{aligned}$$

Thus, equation (1) and $x=5$ are equivalent.

Given,
$$log(7) = log (x^2)rightarrow(1)$$

In order to check whether the equation (1) and $7 = x^2$ are equivalent or not, we will simplify the equation (1) by taking antilog both sides:
$$begin{aligned}
log(7)&= log(x^2)\
&boxed{7=x^2}
end{aligned}$$

Thus, equation (1) and $7=x^2$ are equivalent.

(a.)

Given,

$$
begin{align*}
log(10) &= log(2x-3) tag{1}
end{align*}
$$

We will solve the left-hand side and the right-hand side.

On cancelling $log$ from both sides we get

$$
begin{align*}
log(10) &= log(2x-3) \
10 &= 2x-3 \
13 &= 2x \
dfrac{13}{2} &= x \
6.5 &= x \
end{align*}
$$

Now put the $x$ in equation (1) to check the solution.\
begin{align*}
log (10) &= log (2 times {6.5} – 3)\
log (10) &= log (13 – 3)\
log (10) &= log (10)\
intertext {Hence, the equations are equivalent.}\
end{align*}

(b.)

Given,

$$
begin{align*}
log(25) &= log(4x^{2}-5x-50) tag{1}
end{align*}
$$

We will solve the left-hand side and the right-hand side.

On cancelling $log$ from both sides we get\
begin{align*}
log(25) &= log(4x^{2}-5x-50) \
25 &= 4x^{2}-5x-50 \
4x^{2} – 5x – 50 – 25 &= 0 \
4x^{2} – 5x -75 &= 0 \
intertext {Using the splitting middle term for solving the equation}.\
4x^{2} -20x + 15x – 75 &= 0\
4x (x-5) + 15 (x-5) &= 0 \
(x-5)(4x+15)&=0\
intertext {By solving this, we get the two values of x.}\
end{align*}

$$
begin{align*}
x – 5 &= 0\
fbox {x = 5}\
4x + 15 &= 0\
4x &= -15\
x &= – dfrac{15}{4}\
fbox {x = -3.75}\
end{align*}
$$

put $x = 5$ in equation (1),\
begin{align*}
log(25) &= log (4(5)^{2} – 5(5) – 50))\
log(25) &= log (100 – 25 – 50))\
log(25) &= log (100 – 75)\
log(25) &= log (25)\
intertext {Hence, the equations are equivalent.}\\
intertext {Now, Put x = -3.75}\
log(25) &= log (4(-3.75)^{2} – 5(-3.75)-50))\
log(25) &= log (56.25 + 18.75 – 50))\
log(25) &= log (25)\
intertext {Hence, the equations are equivalent.}\
end{align*}

$(a.) x = 6.5$

$(b.) x = 5 ; -3.75$

$a$.
$$begin{aligned}
sqrt{x+18}&= x-2rightarrow(1)\
end{aligned}$$

Squaring both sides in equation (1):
$$begin{aligned}
x+18&= x^2+4-2x\
0&= x^2-2x-18x+4\
0&= x^2-20x+4rightarrow(2)\
end{aligned}$$

Solving further using discriminant:
$$begin{aligned}
x&= dfrac{(20)pmsqrt{(20)^2-4(1)(4)}}{2(1)}\\
&=dfrac{20pmsqrt{400-16}}{2}\\
&= dfrac{20pmsqrt{384}}{2}\\
&= dfrac{20pm19.6}{2}\
end{aligned}$$
Thus,
$$begin{aligned}
x&=dfrac{20+19.6}{2}\\
x&= dfrac{39.6}{2}\\
&boxed{x=19.8}\
end{aligned}$$
or,
$$begin{aligned}
x&=dfrac{20-19.6}{2}\\
x&= dfrac{0.4}{2}\\
&boxed{x=0.2}\
end{aligned}$$

$b$.
$$begin{aligned}
|x+3|&= -2x+9rightarrow(1)\
end{aligned}$$

Re-writing equation (1):
$$begin{aligned}
pm (x+3)&= -2x+9rightarrow(2)\
end{aligned}$$

Solving equations formed in $(2)$:
$$begin{aligned}
x+2x&= 9-3\
3x&=6\\
x&=dfrac{6}{3}\\
&boxed{x=2}\
end{aligned}$$
Also,
$$begin{aligned}
-x-3&=-2x+9\
-x+2x&= 9+3\
x&= 12\
&boxed{x=12}\
end{aligned}$$

$c$.
$$begin{aligned}
y + 3&= 2x^2-5xrightarrow(1)\
3x+y&=1rightarrow(2)\
end{aligned}$$

From equation (2):
$$begin{aligned}
y& =1 -3xrightarrow(3)\
end{aligned}$$

Put the value of $y$ concluded in equation (3) in equation (1):
$$begin{aligned}
1-3x+3&= 2x^2-5x\
0&= 2x^2-5x+3x-4\
0&= 2x^2-2x-4rightarrow(4)\
end{aligned}$$

Solving further using discriminant:
$$begin{aligned}
x&= dfrac{(2)pmsqrt{(2)^2-4(2)(-4)}}{2(1)}\\
&=dfrac{2pmsqrt{4+32}}{2}\\
&= dfrac{2pmsqrt{36}}{2}\\
&= dfrac{2pm6}{2}\
end{aligned}$$
Thus,
$$begin{aligned}
x&=dfrac{2+6}{2}\\
x&= dfrac{8}{2}\\
&boxed{x=4}\
end{aligned}$$
or,
$$begin{aligned}
x&=dfrac{2-6}{2}\\
x&= dfrac{-4}{2}\\
&boxed{x=-2}\
end{aligned}$$

Solving for $x$:
Put $y = -2 text{or} 4$ in equation (3):
Case 1:
$$begin{aligned}
y&= 1 – 3(4)\
y&= 1-12\
&boxed{y=-11}\
end{aligned}$$
Also,
$$begin{aligned}
y&= 1 – 3(-2)\
y&= 1+6\
&boxed{y=7}\
end{aligned}$$

$d$.
$$begin{aligned}
y&= 2x^2 – 8x+ 7rightarrow(1)\
y&= -2x^2+12x-14rightarrow(2)\
end{aligned}$$

Equating equation (1) and (2) :
$$begin{aligned}
2x^2 – 8x+ 7&= -2x^2+12x-14\
2x^2+2x^2-8x-12x+7+14&=0\
4x^2-20x^2+21&=0rightarrow(3)\
end{aligned}$$

Solving further:
$$begin{aligned}
4x^2-6x-14x+21&=0\
2x(2x-3)-7(2x-3)&=0\
(2x-7)(2x-3)&=0\
end{aligned}$$

Thus,
$$begin{aligned}
2x-7&=0\
2x&=7\
&boxed{x=dfrac{7}{2}}\\
end{aligned}$$
Also,
$$begin{aligned}
2x-3&=0\
2x&= 3\
&boxed{x=dfrac{3}{2}}\
end{aligned}$$

Put $x= 3.5 text{and} 1.5$ in equation (1):
$$begin{aligned}
y&= 2(3.5)^2- 8(3.5) + 7\
y&= 2(12.25)- 28+7\
y&= 24.5-28+7\
&boxed{y=10.5}
end{aligned}$$
Also,

$$begin{aligned}
y&= 2(1.5)^2- 8(1.5) + 7\
y&= 2(2.25)- 12+7\
y&= 4.5-12+7\
&boxed{y=14.5}
end{aligned}$$

Given equations:
$$begin{aligned}
ygeq |x+2|-3rightarrow(1)\
yleq2rightarrow(2)\
end{aligned}$$

$textbf{a.}$

You can use your calculator, but for the purpose of demonstration, we are going to use a graphing calculator on desmos.com.

$textbf{b.}$

Add the control limits. Equations of the lines are $y=150$ and $y=350$. We will also zoom out a bit for a better view.

Draw scatterplot using the given data, and lines $y=150$ and $y=350$ which represent control limits.

$textbf{a.}$

You can use your calculator, but for the purpose of demonstration, we are going to use the graphing calculator on desmos.com.

Add the control limits. Equations of the lines are $y=150$ and $y=350$.

$textbf{b.}$

The process went out of control at hour 15. It is possible that the machine malfunction or the human error is responsible for the issue.

$textbf{a.}$ Create an $x$-bar chart by making a scatterplot and adding the control limits.

$textbf{b.}$ The process went out of control at hour 15. It is possible that the machine malfunction or the human error is responsible for the issue.

$textbf{a.}$

You can use your calculator, but for the purpose of demonstration, we are going to use the graphing calculator on desmos.com. Create the scatterplot.

Add the control limits. Equations of the lines are $y=150$ and $y=350$.

$textbf{b.}$

It can be noticed that mean sizes of ice cubes gravitate towards the upper control limit as hours pass. Engineers probably assessed that the process will go out of control at hour 20.

$textbf{c.}$

The size in the center of the control limits is the mean value of the limits, so

$$
frac{150+350}{2}=250text{ mm}.
$$

$textbf{d.}$

Graph the dotted line $y=250$ representing the centerline.

Since nine points lie above the centerline, the process is out of control.

$textbf{e.}$

For nine points loss of control, I believe that machine malfunction might be a better explanation than human error.

$textbf{a.}$ Draw the scatterplot and add the control limits.

$textbf{b.}$ It was probably assessed that the process would go out of control at hour 20.

$textbf{c.}$ The centerline is at 250 mm cube size.

$textbf{d. }$ Graph the line $y=250$ representing the centerline. The process is out of control.

$textbf{e.}$ Machine malfunction is the likely problem.

Yes, because 20% is within the margin of error: $(13%-10%, 13%+10%)=(3%,20%)$.

Yes, because the specifications are $12pm 0.25=(11.75,12.25)$ and 11.97 is in this interval.

It is given that,

$$
begin{align*}
text{Total number of songs in a playlist} &= 6\
text{Favorable number of her parents wedding song} &= 1\
text{Favorable number of reggae songs} &= 2\
text{Favorable number of jazz songs} &= 3\
end{align*}
$$

begin{align*}
intertext {Let $(E_{1})$ be an event that the song is her parent’s wedding song. Thus, the probability of the song is her parent’s wedding song is mathematically stated below.}\
P(E_{1}) &= dfrac{text{Total numbers of outcomes}}{text{Favorable number of outcomes}}\
P (E_{1}) &= dfrac {1}{6} \
end{align*}

begin{align*}
intertext {Let $(E_{2})$ be an event of reggae song. Thus, the probability of the reggae songs are mathematically stated below.}\
P (E_{2}) &= dfrac {2}{6}\
P(E_{2}) &= dfrac {1}{3}\
end{align*}

begin{align*}
intertext {Let $(E_{3})$ be an event of jazz music. Thus, the probability of jazz music are mathematically stated below.}\
P (E_{3}) &= dfrac {3}{6}\
P(E_{3}) &= dfrac {1}{2}\
end{align*}

$(a.)$

The area model of the given situation is given below.

$(b.)$\
Let $(E^{-1})$ be an event that the song will not be her parent’s wedding song. The probability of the song that will not be her parent’s wedding song is mathematically expressed as follows.\
As we know, the probability of a parent’s wedding song,i.e, \
begin{align*}
P(E_{1}) &= dfrac {1}{6}\
intertext {Now}\
P(E_{1}) + P(E^{-1}) &= 1\
P(E^{-1}) &= 1 – P(E_{1})\
P (text {not her parent’s song}) &= 1 – dfrac{1}{6}\
&= dfrac {5}{6}\
end{align*}

It is given that, the wedding song plays for a mom, or a jazz song plays for dad. Riley’s phone shows a smiling emoticon.

Let $(E_{4})$ be an event that the phone will show a smiling emoticon. Thus, the probability of the phone will show an emoticon is mathematically expressed as follows.

$$
begin{align*}
P(E_{4}) &= dfrac{1}{2} times(text{Probability of wedding song}) + dfrac{1}{2} times(text {Probability of jazz song})\\
&= dfrac{1}{2}times dfrac{1}{6} + dfrac{1}{2} times dfrac{1}{2}\\
&= dfrac{1}{12} + dfrac {1}{4}\\
&= dfrac {1}{3}\\
P(E_{4}) &= 0.33\\
&= 33.3 text{percent}\
end{align*}
$$

$(a.)$ $text {See the explanations}$

$(b.)$ $dfrac {5}{6}$

$(c.)$ $35 text{percent}$

$(d.)$ $75 text {percent}$

It is given that.\
begin{align*}
text {y} &= text {log (x)} tag{1}\
intertext {The logarithum function, $y = log_{b}(x)$, can be shifted $k$ units vertically and $h$ units horizontally with the equation is given below:}\
text {y} &= text {log}_{b} (text{x+h}) + text {k} tag{a} \\
intertext {As comparing the given equation to the logarithum function, we get the value of $b$.}\
text {log}_{b}(text {x}) &= text {log (x)}\
text {b} &= 1\
end{align*}

A linear approximation of $W(x)$ is shown alongside the inverse function $W^{-1}(x)$ obtained by interchanging the x and y coordinates of the points $left{ (-1,0),(0,1),(1,0),(2,-1)right}$.The inverse is not a function since it fails the vertical line test meaning that an element of the domain is mapped to more than one element in the range.

The cockpits of some fighter jets were designed for men between 5’8″ and 6’3″ tall.

As per the U.S. Center for Disease Control average height of a woman in the United States is 63.8 inches with a standard deviation of 3.8 inches. Thus, with respect to women of the U.S. it is given
Average height ($bar x$) = 63.8 inches
Standard Deviation ($sigma$) = 3.8 inches

The cockpit is made for people with the height between
5 feet 8 inches or $(5 times 12) + 8 = 68$ inches and
6 feet 3 inches or $(6 times 12) + 3 = 75$ inches.

We have to find the percentage of women who are not within the optimal height range.

Here, we will use the concept of $z$- score. $z$- score tells us the distance i.e. the standard deviation between the mean and the sample score.
In this case we are given the standard deviation and the mean. The sample scores are the optimal range of heights for which the cockpit is designed.

It is pertinent to note,
$$z=dfrac{x-bar{x}}{sigma}$$

Finding the percentage of women with height less than 68 invhes
$$begin{aligned}
x_a&=68 \
bar{x}& = 63.8 \
sigma &= 3.8\
end{aligned}$$
Calculating the $z$-score,
$$begin{aligned}
z_a&=dfrac{x_a-bar{x}}{sigma} \\
&=dfrac{68-63.8}{3.8} \\
&=dfrac{4.2}{3.8}\\
&=1.10526
end{aligned}$$

Now we will use the normal distribution table,

We get
$x<68 = 0.86548$

This implies,
Percentage of women below the height of 68 inches = $86.548 %$

We can also take help of the graph


Finding the percentage of women with height less than 75 inches
$$begin{aligned}
x_b&=75 \
bar{x}& = 63.8 \
sigma &= 3.8\
end{aligned}$$
Calculating the $z$-score,
$$begin{aligned}
z_b&=dfrac{x_b-bar{x}}{sigma} \\
&=dfrac{75-63.8}{3.8} \\
&=dfrac{11.2}{3.8}\\
&=2.9474
end{aligned}$$

Now we will use the normal distribution table,

We get
$x75 = 1-0.9984=0.0016025$
This implies,
Percentage of women above the height of 75 inches = $0.16025 %$
We can also take help of the graph


We have,
Percentage of women below the height of 68 inches = $86.5 %$
Percentage of women above the height of 75 inches = $0.16 %$

Percentage of women within the optimal range of height i.e. 68 to 75 inches = $100-86.5-0.16 = 13.34 %$

the percentage of women who are not within the optimal height range = $100-13.34 = 86.66 %$

This is represented by the un-shaded region in the graph below

$$86.66 %$$

Given,
$$begin{aligned}
text{ Total case that Walter likes the suit}, n(L)&=0.75\
text{Total case of disliking by Walter}, n(L)^{‘}&= 0.25\
text{Favorable cases of liking}&=7\
text{Favorable cases of dis-liking}&=3\
text{Favorable cases of telling truth}&=1\
end{aligned}$$

Let $E_1$ be an event that Walter says he does not like Bill’s suit. Mathematically it is expressed as,
$$begin{aligned}
P(E_1)&= dfrac{text{Favorable Cases}}{text{Total Cases}}\\
&= dfrac{3}{0.25}\\
&= 12\
end{aligned}$$

Let $E_2$ be an event that Walter says he does like Bill’s suit. Mathematically it is expressed as,
$$begin{aligned}
P(E_2)&= dfrac{text{Favorable Cases}}{text{Total Cases}}\\
&= dfrac{7}{0.75}\\
&= 9.3\
end{aligned}$$

$textbf{a.}$

Create the area model:

$textbf{b.}$

As it can be seen from the area model, she will correctly identify a redhead 4 times out of 100, so we can expect that she will be correct 4 times.

$textbf{c.}$

Lilly will say that the student has red hair if she is correct and the student indeed has red hair, and if she is not correct, and the student does not have red hair. Adding these probability from the area model:

$$
frac{4}{100}+frac{19}{100}=frac{23}{100}.
$$

Out of 100 students, we can expect that Lilly will proclaim 23 of them as redheads.

$textbf{d.}$

We established in $textbf{c.}$ that she will proclaim 23 out of 100 students as redheads, but she is correct in identifying redheads 4 times out of 100. So, the required percentage is

$$
frac{frac{4}{100}}{frac{23}{100}}=frac{4}{23}approx 17%.
$$

$textbf{e.}$

As established in $textbf{d.}$, when she is identifying redheads, she is correct only in $17%$ of the cases, not a great percentage, so it is unlikely that out of Measley’s is a thief.

$textbf{f.}$

This is not a proof. It is merely a probability question, so it is still possible that one of them is a thief, although unlikely.

$textbf{a.}$ Area model.

$textbf{b.}$ 4

$textbf{c.}$ 23

$textbf{d.}$ $17%$

$textbf{e.}$ It is unlikely that Measley is a thief.

$textbf{f.}$ This is not a proof, so a possibility that Measley is a thief still exists.

$textbf{a.}$

Create the area model:

$textbf{b.}$

Out of 100,000 people, 100 are actually sick. Out of 100,000 people who are not sick, but are going to be told so is 999. So, to compare the numbers:

$$
frac{999}{100}approx 10.
$$

The amount of people who are going to be told they are sick and are not is 10 times greater than the number of people who are actually sick.

$textbf{c.}$

The probability of test coming back positive is if the test is correct, and the person is positive, and if the test is not correct and the person is negative. That probability is $frac{99}{100,000}+frac{999}{100,000}=frac{1,098}{100,000}$.

The conditional probability is

$$
frac{frac{999}{100,000}}{frac{1,098}{100,00}}=frac{999}{1,098}approx 91%.
$$

The probability of being false positive is approximately $boldsymbol{91%}$.

$textbf{d.}$

Since the disease is very rare, being proclaimed as HIV positive does not necessarily be the truth. Consecutive testing is recommended. When people volunteer for testing, the percentage of false positives will decrease because much smaller percentage of HIV negative people will submit to the test.

$textbf{e.}$

The test is always $99%$ correct, whether or not the person is HIV positive, so the results are independent.

$textbf{a.}$ Area model.

$textbf{b.}$ There will be approximately 10 times more people that are going to be told they are sick and are actually not than a number of people who are sick.

$textbf{c.}$ $91%$

$textbf{d.}$ Percentage of false positives would reduce.

$textbf{e.}$ Yes.

We ran a simulation in MS Excel. Actually, Jack was the one who won here, but we can conclude that someone will win eventually and they will not be able to play the whole day. Jill actually has greater chances of winning since she has more coins at the beginning. If you want to extend your knowledge on this problem, we encourage you to search and study the $textbf{Gambler’s Ruin}$ problem.

Logarithms are defined as
$$a=b^x Leftrightarrow x=log_ba$$
Where $a,bin R^+$ and $bneq 1$

(a)
$$a=log_b(24)Rightarrow b^a=24$$

(b)
$$3x=log_{2y}(7)Rightarrow (2y)^{3x}=7$$

(c)
$$3y=7^{5x}Rightarrow 5x=log_{2}(3y)$$

(d)
$$4p=(2q)^{6}Rightarrow 6=log_{2q}(4p)$$

Let us consider a real valued function on a domain $~~D subset mathbb R~~$ given by
$$f(x)=frac{2}{7-x},~~~x in D.$$
$(a)~~$ Now notice that denominator of the given function $~~f~~$ is $~~7-x.$
Now if we put $~~7~~$ in the function $~~f(x)~~$the denominator become $~~0~~$ and the function $~~f~$ becomes $~~dfrac{2}0,~~$ which is basically undefined on real line.
So the value of $~~f(7)~~$ cannot be determined.

$(b)~~$ By our assumption $~~D~~$ is the domain of the function $~~f$.
Now notice that
$$f(x)=frac{g(x)}{h(x)},~~~text{ where }~h(x)=7-x,~g(x)=2.$$
Now note that $~~g(x)~~$ is defined on the real line, since $~~g~~$ is a constant function. Also, $~~h(x)~~$ is defined on the real line and $~~h(7)=0.$
Since $~~h(x)~~$ is the denominator of the given function $~~f(x)~~$ and $~~7~~$ is the only zero of $~~f,~~$ $~f~$ cannot be defined on $~~x=7.$
Therefore, the domain $~~D~~$ of the function $~~f~~$ is given by
$$D=mathbb R -{7}.$$

$(c)~~$ Let $~~~S~~$ be the range of the function $~~f.$
Let $~~y~~$ be an element of $~~mathbb R~~$ such that $~~y=f(x).$
Then we have
$$
begin{aligned}
y=f(x) implies & y=frac{2}{7-x}\
implies & y(7-x)=2\
implies & 7y-yx=2\
implies & 7y = yx+2\
implies & 7y-2=xy\
implies & x=frac{7y-2}{y}.
end{aligned}
$$

This follows that if $~~y~~$ be an element of $~~mathbb R~~$ such that $~~f(x)=y,~~$ then we have $~~x=dfrac{7y-2}{y}.$
Now notice that denominator of $~~x~~$ is $~~1+y.$
So, the value of $~~x~~$ can be determined if and only if $~~y neq 0.$

Hence, $~~y in mathbb R~~$ belongs to the range of $~~f,~~$ then $~~y neq 0.$
Hence, the range $~~S~~$ of $~~f~~$ is given by
$$S=mathbb R -{0}.$$
This completes the solution.

Let us consider a function $~~f(x)~~$ given by
$$f(x)=-x+6.$$
Let $~~f^{-1}~~$ be the inverse of the function $~~f.~~$ Then we have
$$left(fcirc f^{-1}right)(x)=left(f^{-1}circ fright)(x)=x.$$

Now notice that
$$
begin{aligned}
y=-x+6 implies & y+x=(-x+6)+x\
implies & y+x=(-x+x)+6\
implies & y+x=0+6\
implies & (y+x)-y=6-y\
implies & (y-y)+x=6-y\
implies & x=6-y.
end{aligned}
$$
Now we claim that $~~f^{-1}(x)=6-x.$

Now notice that
$$
begin{aligned}
left(fcirc f^{-1}right)(x) &=fleft(f^{-1}(x)right)\
&=f(6-x)\
&=-(6-x)+6\
&=-6+x+6\
&=(-6+6)+x\
&=0+x\
&=x,\ \

left(f^{-1}circ fright)(x) &=f^{-1}left(f(x)right)\
&=f^{-1}(-x+6)\
&=6-(-x+6)\
&=6+x-6\
&=(6-6)+x\
&=0+x\
&=x.
end{aligned}
$$

This follows that for all $~~x~~$ we have
$$left(fcirc f^{-1}right)(x)=left(f^{-1}circ fright)(x)=x.$$
Hence, inverse of $f~~$ is defined by
$$f^{-1}(x)=6-x.$$
This completes the solution.

$$f^{-1}(x)=6-x.$$

$2 sqrt {21-x}-sqrt {3x-6}=5$          (Write the equation)

$2 sqrt {21-x}=5+sqrt {3x-6}$

$4(21-x)=25+10sqrt {3x-6}+(3x-6)$          (Square each side)

$84-4x-25-3x+6=10 sqrt {3x-6}$          (Grouping similar terms)

$-7x+65=10 sqrt {3x-6}$

$(-7x+65)^2=100(3x-6)$          (Square each side)

$49x^2-910x+4225=300x-600$

$49x^2-910x-300x+4225+600=0$          (Grouping similar terms)

$49x^2-1210x+4825=0$

Solving the equation using quadratic formula:

$x_{1, 2}=dfrac {-b pm sqrt {b^2-4ac}}{2a}$

$x_{1, 2}=dfrac {1210 pm sqrt {1,464,100-945,700}}{98}$

$x_{1, 2}=dfrac {1210 pm sqrt {518,400}}{98}$

$x_{1}=dfrac {1210-720}{98}$          $x_{2}=dfrac {1210+720}{98}$

$x_{1}=dfrac {1210-720}{98}$          $x_{2}=dfrac {1210+720}{98}$

$x_{1}=5$          $x_{2}=dfrac {965}{49}$

Checking solution:

For $x=5$

$2 sqrt {21-x}-sqrt {3x-6}$

$=2 sqrt {21-5}-sqrt {3(5)-6}$

$=2 sqrt {16}-sqrt {9}$

$=8-3=5$          checkmark

$x=5$ is a real solution.

For $x=dfrac {965}{49}$

$2 sqrt {21-dfrac {965}{49}}-sqrt {3 left(dfrac {965}{49} right)-6}$

$=2 sqrt {dfrac {1029}{49}-dfrac {965}{49}}-sqrt {left(dfrac {2895}{49} right)-left(dfrac {294}{49} right)}$

$=2 sqrt {dfrac {64}{49}}-sqrt {dfrac {2601}{49}}$

$=2 times dfrac {8}{7}-dfrac {51}{7}$

$=dfrac {16}{7}-dfrac {51}{7}$

$=-dfrac {35}{7}=-5$

$x=dfrac {965}{49}$ is an extraneous solution.

The real solution is $x=5$

$textbf{(a)}$ It can be observed in the pattern that:

* $bullet$ Base of each consecutive triangle is increased by 1
* $bullet$ Perpendicular of each consecutive triangle is increased by $sqrt3$
* $bullet$ Hypotenuse of each consecutive triangle is increased by 2

Hence, below are next two triangles:

$textbf{(b)}$ It can be observed in the pattern that:

* $bullet$ Perpendicular of each triangle is $sqrt3 times textrm{base}$.
Hence, perpendicular is $xsqrt3$

* $bullet$ Hypotenuse of each triangle is $2 times textrm{base}$.
Hence, perpendicular is $2x$

(b) Perpendicular=$xsqrt{3}$, Hypotenuse=$2x$

$textbf{a.}$

The difference in proportions is

$$
25% – 15% = 10% = boldsymbol{0.1}.
$$

$textbf{b.}$

$textit{i.}$

She managed to hit $25%$ of 40 putts with her new club which is $textbf{10 putts}$. She also hit $15%$ of 40 putts with her new club which is $textbf{6 putts}$. In total, 16 putts went into the hole.

$textit{ii.}$

Numbers 1 to 16 will represent punts that went into the hole, and other numbers represent punts that did not go into the hole.

$textit{iii.}$

Let’s use MS Excel software to run the simulation. Putts which went into the hole are colored $color{#4257b2}text{blue}$ and those which did not go into the hole are colored $color{#c34632}text{red}$.

We can see that 12 putts went into the hole, which is $frac{12}{80}=15%$, while 64 putts, or $85%$ did not go into the hole. The difference in proportions is $15%-85%=-70%=boldsymbol{-0.7}$.

$textbf{c.}$

The difference of zero means that there is no difference which club is used. Zero is a plausible result since it is in the interval of confidence.

$textbf{d.}$

Since the difference could be negative, we can say that there might not be a true difference between the clubs.

$textbf{a.}$ $0.1$

$textbf{b.}$

$textit{i.}$ 16

$textit{ii.}$ Numbers 1-16 represent hits, while numbers 17-80 represent misses.

$textit{iii.}$ Run the simulation. We got the difference of proportions of $-0.7$.

$textbf{c.}$ Zero means that there is no significant difference between clubs used.

$textbf{d.}$ It’s possible that there is no difference between clubs.

$textbf{a.}$

Create the area model:

$textbf{b.}$

Probability that the alarm is activated is

$$
frac{96}{100,000}+frac{3,996}{100,000}=frac{4,092}{100,000}.
$$

Probability that the alarm is activated, but the event A did not occur is $frac{3,996}{100,000}$.

The required conditional probability is

$$
frac{frac{3,996}{100,000}}{frac{4,092}{100,000}}=frac{3,996}{4,092}approx boldsymbol{97.6%}.
$$

$textbf{c.}$

The accuracy of a detector is the same whether the even occurred or not. Hence, the results are independent.

$textbf{a.}$ Area model.

$textbf{b.}$ $97.6%$

$textbf{c.}$ Yes.

It is given that.

$$begin{aligned}
y &= log_{b} (x)\
end{aligned}$$
As we can see, the graph is decreasing which means, the base is;
0 < b < 1.

From the graph, we notice that,
$$begin{aligned}
-2 < log_{b}(2) < -1\\
log_{b}(b^{-2}) < log_{b}(2) 2 > dfrac{1}{b}\\
1 > 2b^{2} > b\\
b^{2} 0\\
b 0\\
dfrac{1}{2} < b < dfrac{1}{sqrt{2}}\\
0.5 < b < 0.7\\
end{aligned}$$

$$0.5 < b < 0.7$$

$$
5x-1=6+4x
$$

Square both sides and solve for x$;$

$$
x=7
$$

Given Function.
$$begin{aligned}
f(x) &= 2 sqrt{dfrac{x-3}{4}} + 1\
end{aligned}$$

We will replace first f(x) with y.

$$begin{aligned}
y &= 2 sqrt{dfrac{x-3}{4}} + 1\
end{aligned}$$
Next, replace all x’s with y and all y’s with x.
$$begin{aligned}
x &= 2 sqrt{dfrac{y-3}{4}} + 1 \
end{aligned}$$

Now, Solve for $y$. We get
$$begin{aligned}
x -1 &= 2 sqrt{dfrac{y-3}{4}}\
dfrac{x-1}{2} &= dfrac{sqrt{y-3}}{4}\
end{aligned}$$
$$begin{aligned}
(dfrac{x-1}{2})^{2} &= (dfrac{y-3}{4})^{2}\
dfrac{(x-1)^{2}}{4} &= dfrac{y-3}{4}\
y – 3 &= (x-1)^{2}\
y &= (x-1)^{2}\
end{aligned}$$
Finally replace $y$ with $f^{-1}(x)$.
$$begin{aligned}
f^{-1} = (x-1)^{2} + 3\
end{aligned}$$

We graph $f(x)$ and $f^{-1}(x)$ keeping in mind that,
Domain of $f(x)$ is (3, infty),
while the domain of $f^{-1}$ is (1, infty).

$$
Q(t)=Q_{o}r^{t}=110000(1.025)^{t}
$$

We have a geometric series here, with multiplier $r=1+2.5% =1.025$. Here $Q_{o}$ is the initial value value at t=0 and Q(t) is the value for all integer t>0.

$$
Q(10)=110000(1.025)^{10}=$140810
$$

$$
Q(t)=200000=110000(1.025)^{t}
$$

(b) $;$

$$
log_{1.025}left( dfrac{200000}{110000}right)=t
$$

$$
1.025^{t}=dfrac{200000}{110000}
$$

$$
log(1.025^{t})=logleft(dfrac{200000}{110000} right)
$$

$tlog(1.025)=0.011t=0.26$ ; $t=24.21years$

$$
Q(t)=Q_{o}r^{t}=182500(0.95)^{t}
$$

(c) $;$

$$
Q(2)=182500(0.95)^{2}=$164706
$$