Chapter 4.1, Problem 29E

a.

To determine

To write: A linear model that represents the world record for men’s mile as a function of the number of years since $1960$ .

Answer to Problem 29E

The linear model is $y=-\frac{1}{200}x+3.91$.

Explanation of Solution

Given information: In $1960$ , record was $3.91\min .$ while in $1980$ record was $3.81\min .$

Calculation:

Slope intercept form: $y=mx+b$

Let $x$ represent years and $y$ represent the record time.

As $1960$ is the starting year, consider it origin $(x=0)$

  $\Rightarrow$ For $1980$ , $x=20$

  $\therefore$ The points are $(0,3.91)$ and $(20,3.81)$

  $y-\mathrm{int}ercept=3.91$ (when $x=0$ )

Calculating slope,

  $m=\frac{y_2-y_1}{x_2-x_1}$

  $\begin{array}{l}=\frac{3.81-3.91}{20-0}\\ =\frac{-0.1}{20}\\ =-\frac{1}{200}\end{array}$

Substituting derived values to get slope intercept form,$\therefore$Equation is $y=-\frac{1}{200}x+3.91$.

b.

To determine

To estimate: The record time in year $2000$ and $2020$ .

Answer to Problem 29E

Record will be $3.71\min .$ in year $2000$

Record will be $3.61\min .$ in year $2020$ .

Explanation of Solution

Given information: Years to be estimated about are $2000$ and $2020$ .

Calculation:

Using the model derived in part a,

  $y=-\frac{1}{200}x+3.91$

In year $2000$ , $x=40$ ,

  $\begin{array}{l}y=-\frac{1}{200}(40)+3.91\\ y=-\frac{1}{5}+3.91\\ y=3.71\min .\end{array}$

Thus, record will be $3.71\min .$

In year $2020$ , $x=60$ ,

  $\begin{array}{l}y=-\frac{1}{200}(60)+3.91\\ y=-\frac{3}{10}+3.91\\ y=3.61\min .\end{array}$

Thus, record will be $3.61\min .$