The difference between the gas prices in 2013 and 2001 is 3.00
The linear equation of the graph
Start by calculating the slope of the line of best fit using:
[tex]m = \frac{y_2 -y_1}{x_2 - x_1}[/tex]
From the line of best fit, we have:
[tex]m = \frac{3 - 2}{8 - 4}[/tex]
[tex]m = \frac{1}{4}[/tex]
The regression equation is then represented as:
[tex]y = m(x -x_1) + b[/tex]
So, we have:
[tex]y = \frac 14(x -4) + 2[/tex]
[tex]y = \frac 14x -1 + 2[/tex]
[tex]y = \frac 14x +1[/tex]
Express as decimals
[tex]y = 0.25x +1[/tex]
The price of the gas
In 2001, the value of x is 1
So, we have:
[tex]y = 0.25(1) +1 =1.25[/tex]
In 2013, the value of x is 13.
So, we have:
[tex]y = 0.25(13) +1 =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/26137159
