Part A: Annual Rate of Change

To find the annual rate of change, we assume exponential decay of the form:

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

Where:

• $V(t)$ is the value of the car after $t$ years,
• $V_0$ is the initial value of the car,
• $r$ is the annual rate of decrease (which we need to find),
• $t$ is the time in years.

In 1995, $V_0 = 32,000$, and in 2000 ($t = 5$ years), the value is $13,000$.

The equation becomes:

$13,000 = 32,000 \cdot (1 - r)^5$

To solve for $r$, we first divide both sides by 32,000:

$\frac{13,000}{32,000} = (1 - r)^5$

$0.40625 = (1 - r)^5$

Now, take the 5th root of both sides to get rid of the exponent:

$1 - r = \sqrt[5]{0.40625}$

Let me compute the 5th root and solve for $r$.The annual rate of decrease, rounded to 4 decimal places, is 0.1649.

Part B: Percentage Form

To express the rate in percentage form, multiply by 100:

$0.1649 \times 100 = 16.49\%$

Thus, the annual rate of decrease is 16.49%.

Part C: Value of the Car in 2004

To find the value of the car in 2004, we use the same exponential decay model. The number of years between 1995 and 2004 is $t = 9$. The formula is:

$V(9) = 32,000 \cdot (1 - 0.1649)^9$

Let me compute the value of the car in 2004.The value of the car in the year 2004, rounded to the nearest 50 dollars, is $6,300.

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Would you like any further details or clarifications?

Here are 5 related questions you may want to explore:

1. How would the value of the car change if the rate of depreciation were 20% annually?
2. How long would it take for the car's value to drop below $5,000 with the same depreciation rate?
3. What is the half-life of the car's value given the same depreciation rate?
4. How does the exponential depreciation model compare to linear depreciation?
5. How would a sudden increase in the depreciation rate impact the car's value over the next 5 years?

Tip: Exponential decay models are commonly used for depreciation because they better capture the faster value drops earlier in an asset's lifetime.