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Brayden purchased a new car in 1995 for $ 20 , 000 $20,000. The value of the car has been depreciating exponentially at a constant rate. If the value of the car was $ 6 , 800 $6,800 in the year 2000, then what would be the predicted value of the car in the year 2006, to the nearest dollar?

Sagot :

ivycoveredwalls

Answer:

$1863

Step-by-step explanation:

The model for exponential decay (depreciation) is [tex]y=y_0e^{-rt}[/tex].

r is the depreciation rate (which can be found first)

t is the time since the initial purchase

y is the value of the car at time t

[tex]y_0[/tex] is the initial purchase price (t = 0)

Year 0 is 1995.

2000 is year 5.

So, [tex]6800=20000e^{-r\cdot5}\\\\\frac{6800}{20000}=e^{-5r}\\[/tex]

To solve for r take the logarithm of both sides.

[tex]-5r=\ln\left({\frac{6800}{20000}\right)\\\\r=-\frac{1}{5}\ln\left({\frac{6800}{20000}\right) \approx 0.215762[/tex]

Now find the value of the car in 2006, which is year t = 11.

[tex]y=20000e^{-0.215762\cdot11} \approx 1863[/tex]

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