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After a new $28,000 car is driven off the lot, it begins to depreciate at a rate of 18.9% annually. Which function describes the value of the car after t years? C(t)=28,000(0.158)t C(t)=28,000(1.575)t C(t)=28,000(0.811)t C(t)=28,000(0.189)t

2 Answers

3 votes

Answer:

C(t)=28,000(0.811)^t

Explanation:

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User Jmpcm
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Answer: C(t) = 28000(0.811)^t

Explanation:

it begins to depreciate at a rate of 18.9% annually. This means that the rate at which the value is decreasing is exponential. We would apply the formula for exponential growth which is expressed as

y = b(1 - r)^ t

Where

y represents the value of the car after t years.

t represents the number of years.

b represents the initial value of the car.

r represents rate of depreciation.

From the information given,

P = 28000

r = 18.9% = 18.9/100 = 0.189

Therefore

y = 28000(1 - 0.189)^t

y = 28000(0.811)^t

Where C(t) represents y, the function becomes

C(t) = 28000(0.811)^t

User Puyol
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