Chapter 8.9, Problem 10HP

To determine

To calculate: To write an equation for each line of fit and to find if slope and y-intercept of the graphs of the equation the same

Answer to Problem 10HP

Equation of line of fit for State A is $y=0.08x-147.68$

Equation of line of fit for State B is $y=0.02x-27.72$

Explanation of Solution

Given information: Table showing the population of two states for several years

Year	Population (millions)
State A	State B
2001	12.4	12.3
2002	12.5	12.3
2003	12.6	12.3
2004	12.7	12.4
2005	12.7	12.4
2006	12.8	12.4

Formula Used:

Equation of line passing through two points is given as

  $y-y_2=\frac{y_2-y_1}{x_2-x_1}\left(x-x_2\right)$

Slope intercept of line is given as

  $y=mx+c$

Where,

  $m=$ Slope

  $c=$ y-intercept

Calculation:

Data is given as

Year	Population (millions)
State A	State B
2001	12.4	12.3
2002	12.5	12.3
2003	12.6	12.3
2004	12.7	12.4
2005	12.7	12.4
2006	12.8	12.4

Consider State A

Point on line are as follows:

  $\left(2001,12.4\right)$ and $\left(2006,12.8\right)$

Equation of line passing through above points is

  $\begin{array}{l}y-12.4=\frac{12.8-12.4}{2006-2001}\left(x-2001\right)\\ y-12.4=\frac{0.4}{5}\left(x-2001\right)\\ y-12.4=0.08\left(x-2001\right)\\ y-12.4=0.08x-160.08\\ y=0.08x-147.68\end{array}$

Consider State B

Point on line are as follows:

  $\left(2001,12.3\right)$ and $\left(2006,12.4\right)$

Equation of line passing through above points is

  $\begin{array}{l}y-12.3=\frac{12.4-12.3}{2006-2001}\left(x-2001\right)\\ y-12.3=\frac{0.1}{5}\left(x-2001\right)\\ y-12.3=0.02\left(x-2001\right)\\ y-12.3=0.02x-40.02\\ y=0.02x-27.72\end{array}$

Slope of the equation remains same, but y-intercept changes

Conclusion:

Hence,

Equation of line of fit for State A is $y=0.08x-147.68$

Equation of line of fit for State B is $y=0.02x-27.72$