A car was valued at $26,000 in the year 1995. The value depreciated to $13,000 by the year 2002.

A) What was the annual rate of change between 1995 and 2002?
 = 
 Round the rate of decrease to 4 decimal places.

B) What is the correct answer to part A written in percentage form?
 =  %.

C) Assume that the car value continues to drop by the same percentage. What will the value be in the year 2006 ?
value = $ 
 Round to the nearest 50 dollars.

To solve this problem, let's break down each part.

### Part A: Annual Rate of Change

The value of the car depreciated from $26,000 in 1995 to $13,000 in 2002. This period spans 7 years.

The annual rate of change can be calculated using the formula for exponential decay:
\[
V = V_0 \times (1 - r)^t
\]
where:
- $ V $ is the final value ($13,000),
- $ V_0 $ is the initial value ($26,000),
- $ r $ is the annual rate of depreciation, and
- $ t $ is the number of years (7).

Rearrange to solve for $ r $:
\[
\frac{V}{V_0} = (1 - r)^t
\]
\[
\frac{13,000}{26,000} = (1 - r)^7
\]
\[
0.5 = (1 - r)^7
\]
Now, take the seventh root of both sides to isolate $ 1 - r $:
\[
1 - r = (0.5)^{\frac{1}{7}}
\]
\[
r = 1 - (0.5)^{\frac{1}{7}}
\]
Calculating $ r $ to 4 decimal places:

\[
r \approx 0.1041
\]

So, the annual rate of decrease is approximately **0.1041**.

### Part B: Annual Rate of Decrease in Percentage Form

To express this rate as a percentage:
\[
r \times 100 = 0.1041 \times 100 = 10.41\%
\]

Thus, the annual rate of decrease is **10.41%**.

### Part C: Projected Value in 2006

To find the value of the car in 2006, which is 4 years after 2002 (a total of 11 years from 1995), we use the same formula:
\[
V = V_0 \times (1 - r)^{11}
\]
Substitute the values:
\[
V = 26,000 \times (1 - 0.1041)^{11}
\]

Calculating $ V $ and rounding to the nearest $50:

\[
V \approx 7,500
\]

So, the projected value of the car in 2006 is approximately **$7,500**.

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### Summary of Answers:
- **A)** The annual rate of change is **0.1041**.
- **B)** The annual rate of decrease in percentage form is **10.41%**.
- **C)** The value of the car in 2006 is approximately **$7,500**.

Would you like further details on any part? 

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Here are 5 related questions to deepen understanding:
1. What formula would apply if the car's value was increasing instead?
2. How would the rate change if the depreciation occurred over 10 years instead of 7?
3. How can you confirm exponential decay versus linear depreciation?
4. How does compound interest differ from depreciation in this context?
5. What would the car’s value be after 15 years with the same rate?

**Tip:** Always double-check rounding requirements when presenting final answers.

A car was valued at $26,000 in the year 1995. The value depreciated to $13,000 by the year 2002.

A) What was the annual rate of change between 1995 and 2002? Round the rate of decrease to 4 decimal places.

B) What is the correct answer to part A written in percentage form?

C) Assume that the car value continues to drop by the same percentage. What will the value be in the year 2006? Round to the nearest 50 dollars.

Learn how to calculate the annual rate of depreciation for a car valued at $26,000 in 1995, which dropped to $13,000 by 2002. This example uses the exponential decay formula to determine the annual percentage decrease and projects future values, rounding to the nearest $50. Ideal for students in Grades 10-12.

Math Problem Statement

A) What was the annual rate of change between 1995 and 2002?

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