A careful analysis of the cost of operating an automobile was conducted by a firm. The following model was developed: ModifyingAbove y with caretequals 4,000plus0.20x , where ModifyingAbove y with caret is the annual cost and x is the miles driven. a) If the car is driven 15,000 miles this year, the forecasted cost of operating this automobileequals nothing (enter your response as a whole number). b) If the car is driven 25,000 miles this year, the forecasted cost of operating this automobileequals nothing (enter your response as a whole number).

a. 7,000

b. 9,000

Explanation:

The model developed is correctly stated as follows:

y = 4,000 + 0.2x ... (1)

a) If the car is driven 15,000 miles this year, the forecasted cost of operating this automobile ls (enter your response as a whole number)

When x = 15,000, we substitute this into equation (1) and calculate y as follows:

y = 4,000 + 0.2 (15,000) = 4,000 + 3,000 = 7,000

Therefore, the forecasted cost of operating this automobile ls $7,000 when the car is driven 15,000 miles this year.

b) If the car is driven 25,000 miles this year, the forecasted cost of operating this automobile is (enter your response as a whole number).

When x = 15,000, we substitute this into equation (1) and calculate y as follows:

y = 4,000 + 0.2 (25,000) = 4,000 + 5,000 = 9,000

Therefore, the forecasted cost of operating this automobile ls $9,000 when the car is driven 25,000 miles this year.

Comparing the answers obtained in (a) and (b) above, we can see that the higher the number of miles the car is driven, the higher is the operating cost. This implies that there is a positive relationship between the number of miles the car is driven and its operating cost.