Chapter 1, Problem 53RE

a.

To determine

The scatter plot of the given data.

Year	Population ( in thousands)
2012	19321
2013	19533
2014	19893
2015	20271
2016	20612
2017	20984

Answer to Problem 53RE

The obtained graph is defined as,

  

Explanation of Solution

Given Information:

The table is defined as,

Year	Population ( in thousands)
2012	19321
2013	19533
2014	19893
2015	20271
2016	20612
2017	20984

Consider the given table,

Year	Population ( in thousands)
2012	19321
2013	19533
2014	19893
2015	20271
2016	20612
2017	20984

Draw the points on the plane to make the scatter plot.

  

Therefore, the above graph is required scatter plot.

b.

To determine

To calculate: The slope of the secant line $P{Q}_i$ for $i=1,2,3$ .

Answer to Problem 53RE

The slopes are $P{Q}_1=212,P{Q}_2=430.334$ , and $P{Q}_3=\text{415}\text{.75}$ .

And, the function is not discontinues

Explanation of Solution

Given information:

The table is defined as,

Year	Population ( in thousands)
2012	19321
2013	19533
2014	19893
2015	20271
2016	20612
2017	20984

Formula Used:

If $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are two points. Then the slope is defined as,

  $m=\frac{y_2-y_1}{x_2-x_1}$

Calculation:

Consider the given table,

The initial point $P$ is $\left(2012,19321\right)$ and the point $Q$ is $\left(2013,19533\right)$ .

Now, find the slope of the point $P{Q}_1$ .

  $\begin{array}{c}P{Q}_1=\frac{19533-19321}{2013-2012}\\ =\frac{212}{1}\\ =212\end{array}$

Now, find the slope of the point $P{Q}_2$ .

The $P$ is $\left(2012,19321\right)$ and the point $Q$ is $\left(2016,20612\right)$ .

  $\begin{array}{c}P{Q}_2=\frac{20612-19321}{2016-2013}\\ =\frac{1291}{3}\\ =430.334\end{array}$

Now, find the slope of the point $P{Q}_3$ .

The $P$ is $\left(2012,19321\right)$ and the point $Q$ is $\left(2017,20984\right)$ .

  $\begin{array}{c}P{Q}_3=\frac{20984-19321}{2017-2013}\\ =\frac{1663}{4}\\ =\text{415}\text{.75}\end{array}$

Therefore, the obtained slopes are $P{Q}_1=212,P{Q}_2=430.334$ , and $P{Q}_3=\text{415}\text{.75}$ .

c.

To determine

To calculate: The slope of the secant line $P{Q}_i$ for $i=1,2,3$ .

Answer to Problem 53RE

The slopes are $P{Q}_1=212,P{Q}_2=430.334$ , and $P{Q}_3=\text{415}\text{.75}$ .

Explanation of Solution

Given information:

The table is defined as,

Year	Population ( in thousands)
2012	19321
2013	19533
2014	19893
2015	20271
2016	20612
2017	20984

Formula Used:

If $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are two points. Then the slope is defined as,

  $m=\frac{y_2-y_1}{x_2-x_1}$

Calculation:

Consider the given table,

The initial point $P$ is $\left(2012,19321\right)$ and the point $Q$ is $\left(2013,19533\right)$ .

Now, find the slope of the point $P{Q}_1$ .

  $\begin{array}{c}P{Q}_1=\frac{19533-19321}{2013-2012}\\ =\frac{212}{1}\\ =212\end{array}$

Now, find the slope of the point $P{Q}_2$ .

The $P$ is $\left(2013,19533\right)$ and the point $Q$ is $\left(2014,19893\right)$ .

  $\begin{array}{c}P{Q}_2=\frac{19893-19533}{2014-2013}\\ =\frac{360}{3}\\ =360\end{array}$

Now, find the slope of the point $P{Q}_3$ .

The $P$ is $\left(2014,19893\right)$ and the point $Q$ is $\left(2015,20271\right)$ .

  $\begin{array}{c}P{Q}_3=\frac{20271-19893}{2015-2014}\\ =\frac{378}{1}\\ =\text{378}\end{array}$

Now, find the slope of the point $P{Q}_4$ .

The $P$ is $\left(2015,20271\right)$ and the point $Q$ is $\left(2016,20612\right)$ .

  $\begin{array}{c}P{Q}_4=\frac{20612-20271}{2016-2015}\\ =\frac{341}{1}\\ =341\end{array}$

Now, find the slope of the point $P{Q}_5$ .

The $P$ is $\left(2016,20612\right)$ and the point $Q$ is $\left(2017,20984\right)$ .

  $\begin{array}{c}P{Q}_5=\frac{20984-20612}{2017-2016}\\ =\frac{372}{1}\\ =372\end{array}$

Now, add all the averages and divide by number of average.

  $\begin{array}{c}{\text{Slope}}_{\text{ave}}=\frac{212+360+378+341+372}{5}\\ =\frac{212+360+378+341+372}{5}\\ =332.6\end{array}$

Therefore, the obtained average is $332.6$ .

d.

To determine

To calculate: The instantaneous slope at 1 July 2013.

Answer to Problem 53RE

The slopes are $P{Q}_1=212,P{Q}_2=430.334$ , and $P{Q}_3=\text{415}\text{.75}$ .

Explanation of Solution

Given information:

The table is defined as,

Year	Population ( in thousands)
2012	19321
2013	19533
2014	19893
2015	20271
2016	20612
2017	20984

Formula Used:

If $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are two points. Then the slope is defined as,

  $m=\frac{y_2-y_1}{x_2-x_1}$

Calculation:

Consider the given table,

The initial point $P$ is $\left(2012,19321\right)$ and the point $Q$ is $\left(2013,19533\right)$ .

Now, find the slope at 1 July 2013.

  $\begin{array}{c}\text{Slope}=\frac{19533-19321}{2013-2012}\\ =\frac{212}{1}\\ =212\end{array}$

Therefore, the obtained slope is 212 thousand in year.

e.

To determine

The reason that why linear growth may or may not be a bad assumption over longer periods of time.

Answer to Problem 53RE

The linear asuumation is incorrect.

Explanation of Solution

Given information:

The table is defined as,

Year	Population ( in thousands)
2012	19321
2013	19533
2014	19893
2015	20271
2016	20612
2017	20984

Consider the given table,

As it can be observed that the slope between all the years is not same as calculated in previous part. For each year the slope is different this implies that the growth of the population in each year is different. The slope is increasing in each year.

And for the linear growth the slope should be same in each year. So it is not possible to calculate the correct estimates of the population in 2020 by using the linear equation. Thus, the linear model may be a bad assumption over the longer periods of time to calculate the population in any year.

Therefore, the linear model is incorrect.