Solve the problem.

-The average value of a certain type of automobile was $15,900 in 1991 and depreciated to $6900 in 1994. Let y be the average value of the automobile in the year x, where x = 0 represents 1991.
Write a linear equation that models the value of the automobile (y) in terms of the year x.
A) $y = - \frac { 1 } { 3000 } x - 6900$
B) $y = - 3000 x + 6900$
C) $\mathrm { y } = - 3000 \mathrm { x } + 15,900$
D) $y = - 3000 x - 2100$

Question

Solve the problem.

-The average value of a certain type of automobile was $15,900 in 1991 and depreciated to $6900 in 1994. Let y be the average value of the automobile in the year x, where x = 0 represents 1991.
Write a linear equation that models the value of the automobile (y) in terms of the year x. 
A) $y = - \frac { 1 } { 3000 } x - 6900$
B) $y = - 3000 x + 6900$
C) $\mathrm { y } = - 3000 \mathrm { x } + 15,900$
D) $y = - 3000 x - 2100$

Quizplus · Accepted Answer

We can use the two points (0, 15900) and (3, 6900) to find the equation of the line. Using the slope formula, we get:

m = (6900 - 15900)/(3 - 0) = -3000

Then, using the point-slope form, we get:

y - 15900 = -3000(x - 0)

Simplifying, we get:

y = -3000x + 15900

Therefore, option C is the correct choice.

Solve the Problem y=−13000x−6900y = - \frac { 1 } { 3000 } x - 6900y=−30001​x−6900

Solve the Problem $y = - \frac { 1 } { 3000 } x - 6900$