Chapter ISG, Problem 1.7.2.1P

To determine

To write : and graph a linear function f(x) that models the J’s 10 km run times and explain.

Answer to Problem 1.7.2.1P

The equation for the linear function is $f(x)=-0.3x+\mathrm{44.83.}$

Explanation of Solution

Given information: J is training to run a marathon, which is a distance of 26.2 miles. He

trains by running at the same pace as he would run in a marathon, but for shorter distances. The table shows J‘s 10 km run times.

10 km Runs
Training Day	Time (min)
0	45
5	43
10	42

Formula used:

The equation for the linear function is given below.

  $\begin{array}{c}y=ax+b.\\ \text{Where},a=\frac{n{\displaystyle \sum xy-({\displaystyle \sum x)({\displaystyle \sum y)}}}}{n{\displaystyle \sum x^2-({\displaystyle \sum x{)}^2}}}.\\ b=\frac{{\displaystyle \sum y-a{\displaystyle \sum x}}}{n}.\\ \text{Where, }\text{n=number}\text{of}\text{observations}.\end{array}$

Calculation:

Using the graphing calculator the equation for the linear function on above table data is:

  $\begin{array}{l}\text{Sum}\text{of}x=15.\\ \text{Sum}\text{}\text{of}y=130.\\ \text{Mean}X=5.\\ \text{Mean}Y=\mathrm{43.333.}\\ \text{Sum}\text{of}\text{squares}(S{S}_x)=50.\\ \text{Sum}\text{of}\text{products}(SP)=-15.\\ \text{Linear}\text{equation}y=ax+b.\\ a=SP/S{S}_x=-15/50=-\mathrm{0.3.}\\ b=M_y-a{M}_x=43.33-(-0.3\times 5)=\mathrm{44.83333.}\\ y=-0.3x+\mathrm{44.83.}\end{array}$

The equation for the linear function is $f(x)=-0.3x+\mathrm{44.83.}$

The graph of the function $f(x)=-0.3x+\mathrm{44.83.}$ is given below.