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ella purchased a new car in 2000 for $27,600. the value of the car has been depreciating exponentially at a constant rate. if the value of the car was $8,300 in the year 2004, then what would be the predicted value of the car in the year 2009, to the nearest dollar?

User Cepriego
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Final answer:

To find the predicted value of the car in the year 2009, we can use the formula for exponential decay. After solving the equation, the predicted value of the car in the year 2009, to the nearest dollar, would be $6,274.

Step-by-step explanation:

To find the predicted value of the car in the year 2009, we can use the formula for exponential decay:

V = V0 · e^(k · t)

where V is the current value, V0 is the initial value, e is the constant 2.71828 (approximately), k is the decay constant, and t is the time in years.

We know that V0 = $27,600 and V = $8,300. We also know that the year difference from 2000 to 2004 is 4 years, so t = 4. Plugging these values into the formula, we can solve for k:

$8,300 = $27,600 · e^(k · 4)

Simplifying the equation:

e^(k · 4) = $8,300/$27,600

e^(k · 4) ≈ 0.3004

Now, we can solve for k by taking the natural logarithm of both sides:

ln(e^(k · 4)) = ln(0.3004)

k · 4 = ln(0.3004)

k ≈ ln(0.3004)/4

k ≈ -0.3515

Now that we have the value of k, we can find the predicted value of the car in the year 2009. The year difference from 2004 to 2009 is 5 years, so t = 5:

V = $27,600 · e^(-0.3515 · 5)

V ≈ $6,274

Therefore, the predicted value of the car in the year 2009, to the nearest dollar, would be $6,274.

User Siva Prakash
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