Question

Blake purchased a new car in 1990 for \$15,200$15,200. The value of the car has been depreciating exponentially at a constant rate. If the value of the car was \$10,000$10,000 in the year 1992, then what would be the predicted value of the car in the year 2001, to the nearest dollar?

Answer 1

To find the predicted value of Blake's car in 2001, we use the exponential decay formula and the values from 1990 and 1992 to solve for the decay constant. Once we have the decay constant, we calculate the value for the year 2001 and round to the nearest dollar.

Step-by-step explanation:

To determine the predicted value of Blake's car in the year 2001, we must first understand the exponential decay formula, which is commonly represented as V(t) = V_0 * e(-kt), where V(t) is the value at time t, V_0 is the initial value, k is the decay constant, and e is Euler's number.

Given the initial value V_0 = $15,200 in 1990 and the value V(2) = $10,000 in 1992, we can solve for k by setting up the equation $10,000 = $15,200 * e(-2k) and solving for k. Once we have k, we can then predict the value in 2001 by plugging t = 11 into the equation V(t) = $15,200 * e(-11k).

Let's go through the calculation:

1. Solve for k: $10,000 = $15,200 * e(-2k), which simplifies to e(-2k) = 10,000/15,200.
2. Find the natural logarithm of both sides to get -2k = ln(10,000/15,200) and solve for k.
3. Use the value of k to evaluate V(11) = $15,200 * e(-11k) for t = 11, which corresponds to the year 2001.

To find the value to the nearest dollar, we round the result of our calculation to the nearest whole number.

Answer 2

$3,000

Step-by-step explanation:

To find the predicted value of the car in 2001, we need to use the exponential decay formula:

V(t) = V0 * e^(-kt)

where V0 is the initial value of the car ($15,200), t is the time in years since 1990, and k is the exponential decay constant. We can find k using the value of the car in 1992:

V(2) = $10,000 = $15,200 * e^(-2k)

Solving for k, we get:

k = -ln(0.654) / 2

Now that we have k, we can find the value of the car in 2001:

V(11) = $15,200 * e^(-11k)

To the nearest dollar, the predicted value of the car in 2001 is $$3,000$.