Question

A delivery van valued at $25,000 today is depreciating in value exponentially at a rate of 12% each year. Which expression represents its expected value in 5 years?

Answer 1

To model the exponential depreciation of the delivery van's value, you can use the formula for exponential decay:

\[
V(t) = V_0 \cdot e^{-kt}
\]

where:
- \( V(t) \) is the value after time \( t \),
- \( V_0 \) is the initial value,
- \( k \) is the rate of decay, and
- \( t \) is the time in years.

In this case, the initial value \( V_0 \) is $25,000, the decay rate \( k \) is 0.12 (12%), and \( t \) is 5 years.

However, since we're dealing with a straightforward exponential decay that can also be expressed as a fraction rather than using Euler's number \( e \), we typically use the following form:

\[
V(t) = V_0 \cdot (1 - r)^t
\]

where \( r \) is the rate of decay. For this scenario:

\[
V(t) = 25000 \cdot (1 - 0.12)^t
\]

Now, we calculate \( (1 - 0.12) \):

\[
1 - 0.12 = 0.88
\]

Now we substitute this back into the formula:

\[
V(t) = 25000 \cdot (0.88)^t
\]

To find the expected value of the van in 5 years, we set \( t = 5 \):

\[
V(5) = 25000 \cdot (0.88)^5
\]

Thus, the expression that represents the expected value of the delivery van in 5 years is:

\[
V(5) = 25000 \cdot (0.88)^5
\]