All Solutions
Page 297: Closure Activity
This is an arithmetic sequence as its each term follows $t(n+1)=t(n)+4$ with $t(0)=-7$
$textbf{(b)}$ $-64 , -16 , -4 , – 1, cdots$
This is a geometric sequence as each term can be represented by the general term $-64cdot 4^{n-1}$
$textbf{(c)}$ $1,0,1,4,9,cdots$
This is neither arithmetic nor Geometric as it doesn’t follows any of the general features of those sequences.
$textbf{(d)}$ $0,2,4, cdots$
This is an arithmetic sequence as its each term follows $t(n+1)=t(n)+2$ with $t(0)=0$
This also looks like a geometric sequence with common ratio$=2$ but a geometric sequence cannot have a term as a 0 as it will require $n$ to be close to negative infinity to attain a value 0.
$textbf{(b)}$ Geometric sequence
$textbf{(c)}$ Neither
$textbf{(d)}$ Arithmetic sequence
The general term of an arithmetic progression is given by
$$
t(n)=t(1)+(n-1)d
$$
where, $t(n)$ represents the $n^text{th}$ term of the sequence , $t(1)$ represents the first term of the sequence, $d$ represents the common difference (sequence generator) of the sequence and $n$ represents the term number.
We have $d=-3$ and $t(1)=4$, putting these values into the general equation we can have our final equation as
$$
begin{align*}
t(n)&=4+(n-1)cdot -3\
t(n)&=4-3n+3\
&boxed{t(n)=-3n+7}
end{align*}
$$
$$
r=dfrac{t(2)}{t(1)}=dfrac{7.2}{6}=1.2
$$
$$
r=dfrac{t(3)}{t(2)}=dfrac{8.64}{7.2}=1.2
$$
We can see that the ration between any two consecutive term is $1.2$ therefore, this sequence is geometric with the common ratio $r=1.2$
The general term of an geometric sequence is given by
$$
t(n)=t(1)(r)^{n-1}
$$
where, $t(n)$ represents the $n^text{th}$ term of the sequence , $t(1)$ represents the first term of the sequence, $r$ represents the common ratio (sequence generator) of the sequence and $n$ represents the term number.
We have $r=1.2$ and $t(1)=6$ , after putting these values into the general equation we have the final equation as
$$
boxed{t(n)=6(1.2)^{n-1}}
$$
$textbf{(b)}$ $t(n)=6(1.2)^{n-1}$
{color{#4257b2}text{a)}}
$$
$$
begin{align*}
3left(2x-1right)+12=4x-3& {;}quad : 6x+9=4x-3tag{Expand}\
&6x+9-9=4x-3-9tag{Subtract 9 from both sides} \\
&6x=4x-12tag{Simplify} \
&6x-4x=4x-12-4xtag{Subtract 4x from both sides} \\
&2x=-12tag{Simplify} \
&frac{2x}{2}=frac{-12}{2}tag{Divide both sides by 2} \
&{color{#c34632}x=-6}
end{align*}
$$
{color{#4257b2}text{b)}}
$$
$$
begin{align*}
frac{3x}{7}+frac{2}{7}=2& {;}quad : frac{3x}{7}+frac{2}{7}-frac{2}{7}=2-frac{2}{7}tag{Subtract 2/7 from both sides}\
&3x=frac{2cdot :7-2}{7}tag{Simplify} \
&frac{7cdot :3x}{7}=frac{12cdot :7}{7}tag{Multiply both sides by 7} \
&3x=12tag{Simplify} \
&frac{3x}{3}=frac{12}{3}tag{Divide both sides by 3 } \
&{color{#c34632}x=4}
end{align*}
$$
{color{#4257b2}text{c)}}
$$
$$
begin{align*}
frac{x-3}{x}=frac{3}{5}& {;}quad :left(x-3right)cdot :5=xcdot :3 tag{Multiply}\
&5x-15=xcdot :3tag{Simplify} \
&5x-15+15=xcdot :3+15tag{Add 15 to both sides} \
&5x=xcdot :3+15tag{Simplify} \
&5x-xcdot :3=xcdot :3+15-xcdot :3tag{Subtract 3*x from both sides} \
&2x=15tag{Simplify} \
&frac{2x}{2}=frac{15}{2}tag{Divide both sides by 2} \
&{color{#c34632}x=frac{15}{2}}
end{align*}
$$
{color{#4257b2}text{d)}}
$$
$$
begin{align*}
4xleft(x-2right)=left(2x+1right)left(2x-3right)& {;}quad : 4x^2-8x=4x^2-4x-3tag{Expand }\
&4x^2-8x-left(4x^2-4xright)=4x^2-4x-3-left(4x^2-4xright)tag{Subtract 4×2-4x from both sides} \\
&-4x=-3tag{Simplify} \
&frac{-4x}{-4}=frac{-3}{-4}tag{Divide both sides by -4} \
&{color{#c34632}x=frac{3}{4}}
end{align*}
$$
color{#4257b2} text{ a) }x=-6
$$
$$
color{#4257b2}text{ b) }x=4
$$
$$
color{#4257b2}text{ c) } x=frac{15}{2}
$$
$$
color{#4257b2} text{ d) }x=frac{3}{4}
$$
Therefore, The equation of a line with slope $4$ and $y$-intercept $-6$ is given by
begin{align*}
y&=(4)x+(-6)\
&boxed{y=4x-6}
end{align*}
We can find coordinates of any point on this line by selecting any value of any one variable first and then finding the other. For example, Let’s put $x=0$ in the equation of line and solving for the value of $y$ as shown below.
begin{align*}
y&=4(0)-6\
y&=-6
end{align*}
Therefore, $(0,-6)$ will be a point on this line. Similarly we can find other points on this line. The table of coordinates for such points is attached below.
begin{center}
begin{tabular}{|p{1cm}|p{1cm}|}
hline
x&y\
hline
0&-6\
hline
1&-2\
hline
2&2\
hline
$dfrac{3}{2}$&0\
hline
end{tabular}
end{center}
Using these points we can draw the graph of the line which is shown below.
$$y-y_1=mcdot (x-x_1)$$
Therefore, the equation of a line of slope $dfrac{3}{2}$ passing through a point $(5,7)$ is given by
begin{align*}
y-(7)&=dfrac{3}{2}cdot (x-5)\
y-7&=dfrac{3}{2}cdot x+dfrac{3}{2}cdot (-5)\
y-7&=dfrac{3}{2}cdot x -dfrac{15}{2}\
y&=dfrac{3}{2}cdot x – dfrac{15}{2}+7\
y&=dfrac{3}{2}cdot x +dfrac{-15+14}{2}\
&boxed{y=dfrac{3}{2}cdot x -dfrac{1}{2}}
end{align*}
We can find coordinates of any point on this line by selecting any value of any one variable first and then finding the other. For example, Let’s put $x=1$ in the equation of line and solving for the value of $y$ as shown below.
begin{align*}
y&=dfrac{3}{2}cdot x -dfrac{1}{2}\
y&=dfrac{3}{2}cdot 1 -dfrac{1}{2}\
y&=dfrac{3-1}{2}\
y&=1
end{align*}
Therefore, $(1,1)$ will be a point on this line. Similarly we can find other points on this line.The table of coordinates for such points is attached below.
begin{center}
begin{tabular}{|p{1cm}|p{1cm}|}
hline
x&y\
hline
1&1\
hline
0&$-dfrac{1}{2}$\
hline
-1&-2\
hline
end{tabular}
end{center}
Using these points we can draw the graph of the line which is shown below.
For defining a line uniquely we need a fixed point also, along with the slope(like in the part (b) ).
The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by
$$y-y_1=mcdot (x-x_1)$$
Therefore, the equation of a line of slope $-dfrac{2}{3}$ passing through a point $(0,0)$ is given by
begin{align*}
y-(0)&=-dfrac{2}{3}cdot (x-0)\
y&=-dfrac{2}{3}cdot x\
&boxed{y=-dfrac{2}{3}cdot x}
end{align*}
The table of coordinates of points on this line is attached below.
begin{center}
begin{tabular}{|p{1cm}|p{1cm}|}
hline
x&y\
hline
0&0\
hline
1&$-dfrac{2}{3}$\
hline
3&-2\
hline
-3&2\
hline
end{tabular}
end{center}
Using these points we can draw the graph of the line which is shown below.
The explicit equation for an arithmetic sequence is similar to the slope-intercept form of a line:
$$
y=mx+b
$$
In this case,
$$
t(n)=mn+t(0)
$$
where $m$ is the common difference.
From the given, $m=-7$ and $t(0)=t(1)-(-7)=17-(-7)=10$ (by working backwards). So, the equation is:
$$
color{#c34632}t(n)=-7n+24
$$
$textbf{b.}$
The explicit equation for a geometric sequence is similar to exponential function:
$$
y=acdot b^x
$$
In this case,
$$
t(n)= t(0)cdot b^n
$$
where $b$ is the common ratio.
From the given, $b=5$ and $t(0)=dfrac{t(1)}{5}=dfrac{3}{5}$ (by working backwards). So, the equation is:
$$
color{#c34632}t(n)= dfrac{3}{5}(5)^n
$$
b. $t(n)=dfrac{3}{5}(5)^n$
$$
begin{align*}
a_1&=17\
a_2&=a_1-7=17-7=10=a_1-7cdot 1\
a_3&=a_2-7=10-7=3=a_1-7cdot 2\
a_4&=a_3-7=3-7=-4=a_1-7cdot 3\
a_n&=a_{n-1}-7=a_1-7cdot (n-1)=17-7cdot (n-1)
end{align*}
$$
Therefore, it represent an arithmetic sequence with first term equal to 17 and common difference equal to $-7$ with the final explicit equation given by
$$
boxed{a_n=17-7cdot (n-1)}
$$
$textbf{(b)}$ Given recursive equation is $t(n+1)=t(n)+6$ with $t(1)=3$
$$
begin{align*}
t(1)&=3\
t(2)&=t(1)+6=3+6=9=t(1)+6cdot 1\
t(3)&=t(2)+6=9+6=15=t(1)+6cdot 2\
t(4)&=t(3)+6=15+6=21=t(1)+6cdot 3\
t(n)&=t(n-1)+6=t(1)+6cdot (n-1)=3+6cdot (n-1)
end{align*}
$$
Therefore, it represent an arithmetic sequence with first term equal to 3 and common difference equal to $+6$ with the final explicit equation given by
$$
boxed{t(n)=3+6cdot (n-1)}
$$
$textbf{(b)}$ $t(n)=3+6cdot (n-1)$
The LSRL equation can be calculated by the procedure explained below.\ First we need to calculate the values for predefined variables $a$ and $b$ which is defined the in terms of $x$ and $y$ which is shown below.\
$$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$$
The table of values is attached below.\\
begin{tabular}{ |p{1.5cm}|p{2cm}|p{3cm}|p{1cm}|p{1cm}|p{1cm}| }
hline
Serial & Temp $(textdegree F)$ $(x)$ & Number of cups $(y)$ &$xy$ & $x^2$ & $y^2$\
hline
1 & 89 & 150 &13350 & 7921 & 22500\
hline
2& 52 & 85 &4420 & 2704& 7225\
hline
3 & 72& 136&9792 & 5184 & 18496\
hline
4 & 65&101&6565 & 4225& 10201\
hline
5 & 72 & 122&8784 & 5184& 14884\
hline
6 & 45& 66&2970 & 2025 & 4356\
hline
7 & 31& 63 &1953 & 961& 3969\
hline
8 & 89& 137 &12193 & 7921 & 18769\
hline
9 & 57& 86 &4902 & 3249 & 7396\
hline
10 & 38& 80 &3040 & 1444 & 6400\
hline
11& 37& 58&2146 & 1369 & 3364\
hline
12& 71& 118 &8378& 5041 & 13924\
hline
$sum$ & 718 & 1202 &78493& 47228 & 131484\
hline
end{tabular}\\
$$
begin{align*}
a&=dfrac{(sum y)(sum x^2)-(sum x)(sum xy)}{n(sum x^2)-(sum x)^2}\
a&=dfrac{(1202)(47228)-(718)(78493)}{12(47228)-(718)^2}\
a&=dfrac{56768056-56357974}{566736-515524}\
a&=dfrac{410082}{51212}\
a&=8.007537\
a&=8tag{On estimating }
end{align*}
$$
$$
begin{align*}
b&=dfrac{n(sum xy)-(sum x)(sum y)}{n(sum x^2)-(sum x)^2}\
b&=dfrac{12(78493)-(718)(1202)}{12(47228)-(718)^2}\
b&=dfrac{941916-863036}{566736-515524}\
b&=dfrac{78880}{51212}\
b&=1.540264\
b&=1.54tag{On estimating }
end{align*}
$$
Therefore, After putting $a=8$ and $b=1.54$, the final equation of LSRL becomes
$$
boxed{y=1.54x+8}
$$
The electronically calculated (via excel) equation of LSRL is shown below. We can verify our result from this scatterplot.
Now, the $R-squared$ value can be calculated by squaring the $r$ as shown below.
$$
R^2=0.9557^2=0.91336
$$
Here, the value of $R^2$ is $0.91336$ which means that approximately 91 percent variation in sales of number of cups can be predicted by the temperature of that day. Rest 9 percent is dependent on the other factors.
$textbf{(d)}$
$bullet$The form of the given data is linear which can be explained with the magnitude of the correlation coefficient $r$. Here, The magnitude of $r$ is $0.9577$ which is very close to 1 and which signifies that the association is linear.
$bullet$The direction of the given data is positive which can be explained with the slope of the LSRL. Here, The slope of the LSRL is $1.54$ which is greater than 0 and which signifies that direction is positive.
$bullet$The strength of the given data is strong which can be explained with the magnitude of the determination coefficient $R-squared$ value. Here, The value of $R^2$ is $0.91$ which is very close to 1 which means that approximately 91 percent variation in sales of number of cups can be predicted by the temperature of that day. Rest 9 percent is dependent on the other factors. Therefore, the association is strong.
$bullet$ The point $(72,136)$ is an outlier for the given data as the magnitude of the value of the residual (17.09) for this data point is largest among all the data points.
$$
begin{align*}
y&=1.54x+8\
y&=1.54(90)+8\
y&=138.6+8\
y&=146.6
end{align*}
$$
Now, as we know that the highest positive residual value is 17.09 which means the sales can vary below this value, so we need to add 17.09 in the predicted value of $y$ to find the number of cups of fruits Mary needs to prepare to make it certain.
$$
146.6+17.09=163.69
$$
Therefore, she should prepare approximately 164 fruit cups to make sure that she does not run out of cups.
$textbf{(b)}$ Yes
$textbf{(c)}$ $r=0.9577$ and $R^2=0.9133$
$textbf{(d)}$ The form is linear, direction is positive and the strength is strong for the given data.
$textbf{(e)}$ Mary should prepare approximately 164 fruit cups to make sure that she does not run out of cups.
Therefore, we can also say that we can reduce the chances of getting cancer by decreasing the coffee intake.
