All Solutions
Page 241: Closure Activity
$$a=dfrac{(sum y)(sum x^2)-(sum x)(sum xy)}{n(sum x^2)-(sum x)^2}$$\
$$b=dfrac{n(sum xy)-(sum x)(sum y)}{n(sum x^2)-(sum x)^2}$$
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{2cm}|p{3cm}|p{1cm}|p{1cm}|p{1cm}| }
hline
Serial & Oaks $(x)$ & Nightingale $(y)$ &$xy$ & $x^2$ & $y^2$\
hline
1 & 8 & 5 &40 & 64 & 25\
hline
2& 13 & 9 &117 & 169 & 81\
hline
3 & 4& 3 &12 & 16 & 9\
hline
4 & 5 &5 &25 & 25 & 25\
hline
5 & 10 & 7 &70 & 100 & 49\
hline
6 & 9 & 7 &63 & 81 & 49\
hline
7 & 4 & 5 &20 & 16 & 25\
hline
$sum$ & 53 & 41 &347& 471 & 263\
hline
end{tabular}
$$
begin{align*}
a&=dfrac{(41)(471)-(53)(347)}{7(471)-(53)^2}\
a&=dfrac{19311-18391}{3297-2809}\
a&=dfrac{920}{488}\
a&=1.885245\
a&=1.9tag{On estimating }
end{align*}
$$
$$
begin{align*}
b&=dfrac{7(347)-(53)(41)}{7(471)-(53)^2}\
b&=dfrac{2429-2173}{3297-2809}\
b&=dfrac{256}{488}\
b&=0.52459\
b&=0.5tag{On estimating }
end{align*}
$$
Therefore, After putting $a=1.9$ and $b=0.5$, the final equation of LSRL becomes
$$
boxed{y=0.5x+1.9}
$$
$y$-intercept in this model represent the number of nightingale in the park if there is not any oak tree in the park. From the LSRL equation there are approximately 1.9 nightingale in the park if there is not even a single oak tree in the park.
$textbf{(b)}$ $y=0.5x+1.9$
$textbf{(c)}$ The $textbf{slope}$ in this context represent the number of nightingales in the park per oak tree present in the park. From the LSRL equation there are 0.5 number of nightingale per oak tree present in the park, which means there is 1 nightingale per two oak trees.
$textbf{$y$-intercept }$in this model represent the number of nightingale in the park if there is not any oak tree in the park. From the LSRL equation there are approximately 1.9 nightingale in the park if there is not even a single oak tree in the park.
$$
begin{align*}
y&=0.5x+1.9\
y&=0.5(6)+1.9\
y&=3+1.9\
y&=4.9
end{align*}
$$
Therefore, Sarah should expect approximately 5 nightingales to find in that particular area with 6 oaks
$textbf{(b)}$ The residual is given by the formula
$$
text{Residual}=text{actual value}-text{predicted value}
$$
The predicted value from the LSRL equation is 5 but the actual value is 4 hence the residual is
$$
text{Residual}=4-5=-1
$$
From the graph we can see that the upper boundary line passes from $(4,5)$ and lower boundary line passes through $(4,3)$.
Now, The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by
$$
y-y_1=mcdot (x-x_1)
$$
$bullet$ Finding equation for upper boundary line
The equation of a line of slope $0.5$ passing through a point $(4,5)$ is given by
$$
begin{align*}
y-5&=0.5cdot (x-4)\
y-5&=0.5cdot x +0.5cdot -4\
y&=0.5cdot x -2+5\
&boxed{y=0.5x +3}
end{align*}
$$
$bullet$ Finding equation for Lower boundary line
The equation of a line of slope $0.5$ passing through a point $(4,3)$ is given by
$$
begin{align*}
y-3&=0.5cdot (x-4)\
y-3&=0.5cdot x +0.5cdot -4\
y&=0.5cdot x -2+3\
&boxed{y=0.5x +1}
end{align*}
$$
$bullet$ For Upper bound
$$
begin{align*}
y&=0.5x+3\
y&=0.5(6)+3\
y&=3+3\
&boxed{y=6}
end{align*}
$$
$bullet$ For Lower bound
$$
begin{align*}
y&=0.5x+1\
y&=0.5(6)+1\
y&=3+1\
&boxed{y=4}
end{align*}
$$
$textbf{(b)}$ Residual$=-1$
$textbf{(c)}$ The upper boundary line equation is $y=0.5x +3$
Lower boundary line equation is $y=0.5x +1$
$textbf{(d)}$ The upper bound is $6$ and lower bound is $4$
Correlation coefficient $(r)$ can be also be calculated manually by the formula given below
$$r=dfrac{n(sum xy)-(sum x)(sum y)}{sqrt{[n(sum x^2)-(sum x)^2][n(sum y^2)-(sum y)^2)]}}$$
Now, the table of calculated variable required in the formula is attached below.\\
begin{tabular}{ |p{1.5cm}|p{2cm}|p{3cm}|p{1cm}|p{1cm}|p{1cm}| }
hline
Serial & Oaks $(x)$ & Nightingale $(y)$ &$xy$ & $x^2$ & $y^2$\
hline
1 & 8 & 5 &40 & 64 & 25\
hline
2& 13 & 9 &117 & 169 & 81\
hline
3 & 4& 3 &12 & 16 & 9\
hline
4 & 5 &5 &25 & 25 & 25\
hline
5 & 10 & 7 &70 & 100 & 49\
hline
6 & 9 & 7 &63 & 81 & 49\
hline
7 & 4 & 5 &20 & 16 & 25\
hline
$sum$ & 53 & 41 &347& 471 & 263\
hline
end{tabular}\\
Therefore, by putting the values from the table, we can calculate the $r$ as shown below.
begin{align*}
r&=dfrac{n(sum xy)-(sum x)(sum y)}{sqrt{[n(sum x^2)-(sum x)^2][n(sum y^2)-(sum y)^2)]}}\
r&=dfrac{7(347)-(53)(41)}{sqrt{[7(471)-(53)^2][7(263)-(41)^2)]}}\
r&=dfrac{2429-2173}{sqrt{[3297-2809][1841-1681]}}\
r&=dfrac{256}{sqrt{[488][160]}}\
r&=dfrac{256}{sqrt{[488][160]}}\
r&=dfrac{256}{sqrt{78080}}\
r&=dfrac{256}{279.42798}\
r&=0.91615735\
r&approx0.92
end{align*}
A positive value of correlation coefficient signifies that there is positive correlation between the number of nightingale and the number of oak trees in the park. which in simple language means that with the increase in the number of oak trees in the park the number of nightingale will increase.
Moreover, more close the value of $r$ to the 1, more close the linear relationship between them will be.
As our $r$ in this case is approximately $0.92$ which is close to 1 , hence we can said that there is approximately linear relationship between them.
Correlation coefficient $(r)$ can be also be calculated manually by the formula given below
$$
r=dfrac{n(sum xy)-(sum x)(sum y)}{sqrt{[n(sum x^2)-(sum x)^2][n(sum y^2)-(sum y)^2)]}}
$$
The table of calculated variable required in the formula is attached inside.
$$
text{Area}=dfrac{1}{2}times text{Base}timestext{Perpendicular height}
$$
Now, from the graph attached below, we can see that it is a right angled triangle.
$Rightarrow$
$$
text{Area}=dfrac{1}{2}times 4times 6=12 text{ Unit}^2
$$
$$
begin{align*}
text{A}(-2,-2) rightarrow text{A’}(2,2)\
text{B}(2,-2) rightarrow text{B’}(2,-2)\
text{C}(-2,4) rightarrow text{C’}(8,2)\
end{align*}
$$
$textbf{(b)}$
$$
begin{align*}
text{A}(-2,-2) rightarrow text{A’}(2,2)\
text{B}(2,-2) rightarrow text{B’}(2,-2)\
text{C}(-2,4) rightarrow text{C’}(8,2)
end{align*}
$$
$textbf{(c)}$ The said transformation can be obtained via translating the $Delta$ABC by 2 units right and 4 units downwards.
Use the exponent rule: $a^{-n}=dfrac{1}{a^n}$
$$
3^{-2}= dfrac{1}{3^{2}}
$$
Evaluate:
$$
= color{#c34632}dfrac{1}{9}
$$
$textbf{b.}$
Use the exponent rule: $(a^{m})^n=a^{mn}$
$$
a^3b^2 (b^{-1})^3= a^3b^2 cdot b^{-3}
$$
Use the exponent rule: $a^{m}cdot a^n=a^{m+n}$
$$
= a^3 b^{-1}
$$
Use the exponent rule: $a^{-n}=dfrac{1}{a^n}$
$$
=color{#c34632}dfrac{a^3}{b}color{white}tag{1}
$$
Use the exponent rule: $dfrac{a^{m}}{a^n}=a^{m-n}$
$$
dfrac{x^2y^3}{x^2y^{-1}}=x^{0}y^{4}
$$
Use the exponent rule: $a^0=1$
$$
=1cdot y^{4}
$$
Simplify:
$$
=color{#c34632} y^{4}
$$
$textbf{d.}$
Group the decimals and group the powers using associative property:
$$
dfrac{4times 10^5}{8times 10^{7}}=dfrac{4}{8}times dfrac{10^{5}}{10^{7}}
$$
Use the exponent rule: $dfrac{a^{m}}{a^n}=a^{m-n}$
$$
=0.5times 10^{-2}
$$
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:
$$
=color{#c34632}5times 10^{-3}
$$
b. $dfrac{a^3}{b}$
c. $y^{4}$
d. $5times 10^{-3}$
{color{#4257b2}text{ a) }}
$$
$$
begin{align*}
&text{This example is solved in the following way}\
&=frac{1}{3^2}\\
&=boxed{{color{#c34632} frac{1}{9} } }
end{align*}
$$
$$
color{#c34632} text{}mathrm{Apply:exponent:rule}:quad :a^{-b}=frac{1}{a^b}
$$
{color{#4257b2}text{ b) }}
$$
$$
begin{align*}
&text{This example is solved in the following way}\
&=b^2left(ab^{-1}right)^3\\
&=b^2left(frac{a}{b}right)^3\
&=b^2frac{a^3}{b^3} \
&=frac{a^3b^2}{b^3} tag{Multiply fractions}\\
&=boxed{{color{#c34632} frac{a^3}{b} } }
end{align*}
$$
$$
color{#c34632} text{}mathrm{Apply:exponent:rule::}a^3b^3=left(abright)^3,:quad mathrm{:assuming:}age :0,:bge :0
$$
$$
color{#c34632} text{}mathrm{Multiply:fractions}:quad :acdot frac{b}{c}=frac{a:cdot :b}{c}
$$
$$
color{#c34632} text{} mathrm{Apply:exponent:rule}:quad frac{x^a}{x^b}:=:frac{1}{x^{b-a}}
$$
$$
text{}
$$
{color{#4257b2}text{ c) }}
$$
$$
begin{align*}
&text{Cancel the common factor}\
&=frac{y^3}{y^{-1}}\\
&=y^{3-left(-1right)} \\
&=boxed{{color{#c34632} y^4 } }
end{align*}
$$
$$
color{#c34632} text{}mathrm{Apply:exponent:rule}:quad frac{x^a}{x^b}:=:x^{a-b}
$$
{color{#4257b2}text{ d) }}
$$
$$
begin{align*}
&text{Cancel the common factor}\
&=frac{10^5}{10^7cdot :2}\\
&=frac{1}{10^{7-5}cdot :2} \
&=frac{1}{10^2cdot :2}\
&=frac{1}{100cdot :2}\\
&=boxed{{color{#c34632} frac{1}{200} } }
end{align*}
$$
$$
color{#c34632} text{}mathrm{Apply:exponent:rule}:quad frac{x^a}{x^b}:=:x^{a-b}
$$
color{#4257b2} text{a)}frac{1}{9}
$$
$$
color{#4257b2} text{b)}frac{a^3}{b}
$$
$$
color{#4257b2} text{c)}y^4
$$
$$
color{#4257b2} text{d)} frac{1}{200}
$$
$$
begin{align*}
(x-1)(x+7)&=(x+1)(x-3)tag{use distributive property}\
x(x+7)-1(x+7)&=x(x-3)+1(x-3)\
xcdot x +xcdot 7-1cdot x-1cdot 7&=xcdot x+xcdot -3+1cdot x +1cdot -3\
x^2+7x-x-7&=x^2-3x+x-3\
x^2+6x-7&=x^2-2x-3 tag{subtract $x^2$ from each side}\
x^2-x^2+6x-7&=x^2-x^2-2x-3\
6x-7&=-2x-3tag{add $2x+7$ to the each side}\
6x+2x+7-7&=-2x+2x+7-3\
8x&=4tag{divide each side by 8}\
dfrac{8x}{8}&=dfrac{4}{8}\
&boxed{x=dfrac{1}{2}}
end{align*}
$$
$textbf{(b)}$
$$
begin{align*}
2x-5(x+4)&=-2(x+3)tag{use distributive property}\
2x-5cdot x-5cdot 4 &=-2cdot x -2cdot 3\
2x-5x-20&=-2x-6\
-3x-20&=-2x-6tag{add $3x+6$ to the each side}\
-3x+3x+6-20&=-2x+3x+6-6\
-14&=xtag{interchange sides}\
x&=-14\
&boxed{x=-14}
end{align*}
$$
$textbf{(b)}$ $x=-14$
$$
begin{align*}
dfrac{x}{2}+dfrac{x}{3}&=2tag{multiply each term by 6 (LCM)}\
dfrac{x}{2}cdot 6+dfrac{x}{3}cdot 6&=2cdot 6\
dfrac{6}{2}cdot x+dfrac{6}{3}cdot x&=2cdot 6\
3cdot x+2cdot x&=12\
3x+2x&=12\
5x&=12tag{divide each side by 5}\
dfrac{5x}{5}&=dfrac{12}{5}\
&boxed{x=dfrac{12}{5}}
end{align*}
$$
$$
textbf{(b)}
$$
$$
begin{align*}
6^{(x+9)}&=36^x\
6^{(x+9)}&=(6^2)^xtag{use $(a^m)^n=a^{mcdot n}$}\
6^{(x+9)}&=6^{2cdot x}\
6^{(x+9)}&=6^{2x}tag{equate powers as base is same}\
Rightarrow\
x+9&=2xtag{subtract $x$ from each side}\
x-x+9&=2x-x\
9&=x\
&boxed{x=9}
end{align*}
$$
$textbf{(c)}$
$$
begin{align*}
(x-13)^3&=8tag{rewrite 8 as $2^3$}\
(x-13)^3&=2^3tag{equate powers as base is same}\
Rightarrow\
x-13&=2tag{add 13 to the each side}\
x-13+13&=2+13\
&boxed{x=15}
end{align*}
$$
$textbf{(b)}$ $x=9$
$textbf{(c)}$ $x=15$
