Question

Zoey purchased a new car in 1992 for \$28,000$28,000. The value of the car has been depreciating exponentially at a constant rate. If the value of the car was \$9,600$9,600 in the year 1996, then what would be the predicted value of the car in the year 1998, to the nearest dollar?

Answer 1

400$
Step-by-step explanation:
In 1992, the price is 28000$ and 9600$ in 1996 so in 4 years the value the car lost is 18400$
Dividing 18400 by 4 years we get 4600 depreciation value per year.
Between 1996 and 1998 there is 2 years
So in 1998 the value of the car will be:
9600 (value in 1996) - 2x4600 = 400$

Answer 2

The predicted value of the car in 1998, based on the calculated exponential depreciation model, is approximately $5,621.18, rounded to the nearest dollar.

We can predict the car's value in 1998:

1. Calculate the depreciation rate:

We know the initial value in 1992 was $28,000 and the value dropped to $9,600 in 1996, a period of 4 years.

This represents a decrease of $18,400 in value over 4 years.

Therefore, the annual depreciation rate is $18,400 / 4 years = $4,600 per year.

2. Model the car's value as an exponential function:

The car's value depreciates exponentially, meaning the rate of decrease remains constant over time.

We can model this using an exponential function of the form:

V(t) = V_0 * e^(-r * t)

Where:

V(t) is the car's value at time t (years since 1992)

V_0 is the initial value ($28,000)

r is the depreciation rate ($4,600 per year)

3. Predict the value in 1998:

We want to find V(6), where t = 6 represents 1998 (6 years after 1992).

Plugging in the values, we get:

V(6) = 28,000 * e^(-4,600 * 6) ≈ $5,621.18

Therefore, the predicted value of the car in 1998 is approximately $5,621.18, rounded to the nearest dollar.