Chapter 1.4, Problem 39E

a.

To determine

To make: A scatter plot of the data in the given table.

Answer to Problem 39E

  

Explanation of Solution

Given information:

Table for U.S. exported wheat

  $\begin{array}{cc}\text{Year}& \text{Exported Wheat }\left(\text{millions of bushels}\right)\\ 2012& 1007\\ 2013& 900\\ 2014& 1155\\ 2015& 842\\ 2016& 754\\ 2017& 1031\end{array}$

Concept used:

Put all the given points in the graphing calculator as x and y co-ordinate and create a scatter plot.

Calculation:

Using graphing calculator, the scatter plot obtained is:

  

b.

To determine

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

Answer to Problem 39E

  $\begin{array}{c}\text{Slope of }P{Q}_1=-55\text{ millionbushels per year}\text{.}\\ \text{Slope of }P{Q}_2=-63.25\text{ millionbushels per year}\text{.}\\ \text{Slope of }P{Q}_3=4.8\text{ millionbushels per year}\end{array}$

Explanation of Solution

Given information:

Table for U.S. exported wheat

  $\begin{array}{cc}\text{Year}& \text{Exported Wheat }\left(\text{millions of bushels}\right)\\ 2012& 1007\\ 2013& 900\\ 2014& 1155\\ 2015& 842\\ 2016& 754\\ 2017& 1031\end{array}$

Concept used:

Let $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ be two points.

Then increment in x , that is, $\Delta x$ is calculated as: $\Delta x=x_2-x_1$ .

Then increment in y , that is, $\Delta y$ is calculated as: $\Delta y=y_2-y_1$ .

The slope between the two points is calculated as: $\frac{\Delta y}{\Delta x}$ .

Calculation:

Let P represents the point corresponding to 2012.

  $Q_1$ the point corresponding to 2015

  $Q_2$ the point corresponding to 2016

  $Q_2$ the point corresponding to 2017

For the slope between $P=\left(2012,1007\right)\text{}\text{and }{Q}_1=\left(2015,842\right)$ is calculated as:

  $\begin{array}{c}\text{Slope of }P{Q}_1=\frac{\Delta y}{\Delta x}\\ =\frac{842-1007}{2015-2012}\\ =-55\text{ millionbushels per year}\end{array}$

For the slope between $P=\left(2012,1007\right)\text{}\text{and }{Q}_2=\left(2016,754\right)$ is calculated as:

  $\begin{array}{c}\text{Slope of }P{Q}_2=\frac{\Delta y}{\Delta x}\\ =\frac{754-1007}{2016-2012}\\ =-\frac{254}{4}\text{ millionbushels per year}\\ =-63.25\text{ millionbushels per year}\end{array}$ .

For the slope between $P=\left(2012,1007\right)\text{}\text{and }{Q}_3=\left(2017,1031\right)$ is calculated as:

  $\begin{array}{c}\text{Slope of }P{Q}_3=\frac{\Delta y}{\Delta x}\\ =\frac{1031-1007}{2017-2012}\\ =\frac{24}{5}\\ =4.8\text{ millionbushels per year}\end{array}$ .

  $\begin{array}{c}\text{Slope of }P{Q}_1=-55\text{ millionbushels per year}\text{.}\\ \text{Slope of }P{Q}_2=-63.25\text{ millionbushels per year}\text{.}\\ \text{Slope of }P{Q}_3=4.8\text{ millionbushels per year}\end{array}$