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9 December, 09:55

A new car is purchased for 16000 dollars. The value of the car depreciates at 6.25% per year. To the nearest tenth of a year, how long will it be until the value of the car is 11500 dollars?

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Answers (2)
  1. 9 December, 10:05
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    Answer: it will take 5.1 years

    Step-by-step explanation:

    We would apply the formula for exponential decay which is expressed as

    A = P (1 - r) ^ t

    Where

    A represents the value of the car after t years.

    t represents the number of years.

    P represents the value of the car.

    r represents rate of decay.

    From the information given,

    A = $11500

    P = $16000

    r = 6.25% = 6.25/100 = 0.0625

    Therefore

    11500 = 16000 (1 - 0.0625) ^t

    11500/16000 ( = 0.9375) ^t

    0.71875 = (0.9375) ^t

    Taking log of both sides to base 10

    Log 0.71875 = tlog 0.9375

    - 0.143 = - 0.028t

    t = - 0.143 / - 0.028

    t = 5.1 years
  2. 9 December, 10:20
    0
    5.1 years

    Step-by-step explanation:

    the formula for fixed rate depreciation is

    Value after t years = initial value (1 - r) ^t

    where r = depreciation rate = 6.25% = 0.0625

    t = length of time in years (we are asked to find this

    Value after t years = given as $11,500

    initial Value = given as $16,000

    Substituting this into equation

    11,500 = 16,000 (1 - 0.0625) ^t

    11,500 = 16,000 (0.9375) ^t (divide both sides by 16,000 and rearrange)

    0.9375^t = 11,500 / 16000

    0.9375^t = 0.71875 (taking log of both sides)

    t log 0.9375 = log 0.71875

    t = log 0.71875 / log 0.9375 (use calculator)

    t = 5.12

    t = 5.1 years (nearest tenth)
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