The least-squares regression equation
y = 8.5 + 69.5x can be used to predict the monthly cost for cell phone service with x phone lines. The list below shows the number of phone lines and the actual cost.

(1, $90)
(2, $150)
(3, $200)
(4, $295)
(5, $350)

Calculate the residuals for 2 and 5 phone lines, to the nearest cent.

The residual for 2 phone lines is $_

The residual for 5 phone lines is $_

The residual is the vertical distance separating the observed point from your expected y-value, or more simply put, it is the difference between the actual y and the predicted y.

How to solve the question?

In the question, we are asked to find the residual for 2 and 5 lines using the least-squares regression equation y = 8.5 + 69.5x and the actual costs given to us.

We know that the residual is the vertical distance separating the observed point from your expected y-value, or more simply put, it is the difference between the actual y and the predicted y.

Thus for 2 phone lines:-

Actual Cost = $150.

Predicted Cost, y = 8.5 + 69.5*2 = 147.5.

Residual = Actual Cost - Predicted Cost = 150 - 147.5 = $2.5.

Thus, the residual for 2 phone lines is $2.5.

Thus for 5 phone lines:-

Actual Cost = $350.

Predicted Cost, y = 8.5 + 69.5*2 = 356.

Residual = Actual Cost - Predicted Cost = 350 - 356 = -$6.

Thus, the residual for 2 phone lines is -$6.

Learn more about the residual in a least-square regression equation at

https://brainly.com/question/20165292

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