The difference between the gas prices in 2013 and 2001 is 3.00
See attachment for the graph of the line of best fit
Start by calculating the slope of the line of best fit using:
[tex]b = \frac{y_2 -y_1}{x_2 -x_1}[/tex]
From the line of best fit, we have:
[tex]b = \frac{3 -2}{8 -4}[/tex]
The regression equation is then represented as:
[tex]\^y = b(\^x - x_1) + y_1[/tex]
So, we have:
[tex]\^y = \frac 14 (\^x - 4) + 2[/tex]
[tex]\^y = \frac 14\^x - 1 + 2[/tex]
[tex]\^y = \frac 14\^x + 1[/tex]
So, the linear regression equation is:
[tex]\^y = \frac 14\^x + 1[/tex]
In 2001, the value of x is 1.
So, we have:
[tex]\^y_1 = \frac 14\times (1) + 1[/tex]
[tex]\^y_1 = 1.25 \\[/tex]
In 2013, the value of x is 13.
So, we have:
[tex]\^y_2 = \frac 14\times (13) + 1[/tex]
[tex]\^y_2 = 4.25[/tex]
So, the difference (d) between the prices is:
[tex]d = 4.25 - 1.25[/tex]
[tex]d = 3.00[/tex]
Hence, the difference between the gas prices in 2013 and 2001 is 3.00
Read more about linear regression at:
https://brainly.com/question/17844286
