To solve these questions, we'll start by determining the annual rate of change between 1995 and 2004 for the value of the car.
A) Annual Rate of Change
- Let \( V_0 \) be the initial value in 1995, which is $35,000.
- Let \( V_t \) be the value in 2004, which is $13,000.
- The time period from 1995 to 2004 is 9 years.
The formula for exponential decay is: \[ V_t = V_0 \times (1 + r)^t \]
Where:
- \( V_t \) is the value at time \( t \)
- \( V_0 \) is the initial value
- \( r \) is the annual rate of change (as a decimal)
- \( t \) is the number of years
We can rearrange the formula to solve for \( r \): \[ r = \left(\frac{V_t}{V_0}\right)^{\frac{1}{t}} - 1 \]
Plugging in the values: \[ r = \left(\frac{13000}{35000}\right)^{\frac{1}{9}} - 1 \]
Calculating \( \frac{13000}{35000} \): \[ \frac{13000}{35000} \approx 0.3714285714 \]
Now calculate \( 0.3714285714^{\frac{1}{9}} \): \[ 0.3714285714^{\frac{1}{9}} \approx 0.7360040824 \]
Subtracting 1 gives us: \[ r \approx 0.7360040824 - 1 \approx -0.2639959176 \]
Rounding to 4 decimal places: \[ r \approx -0.2640 \]
B) Rate in Percentage Form
To convert the decimal to percentage, we multiply by 100: \[ r \approx -0.2640 \times 100 \approx -26.40% \]
C) Value in 2009
Now we need to calculate the value in 2009, which is 5 years after 2004. We'll use the same decay formula:
\[ V_{2009} = V_{2004} \times (1 + r)^t \] Where:
- \( V_{2004} = 13000 \)
- \( r \approx -0.2640 \)
- \( t = 5 \)
Calculating: \[ V_{2009} = 13000 \times (1 - 0.2640)^5 \]
Calculating \( (0.7360)^5 \): \[ (0.7360)^5 \approx 0.2067986836 \]
Now, multiplying: \[ V_{2009} = 13000 \times 0.2067986836 \approx 2688.373886 \]
Rounding to the nearest $50: \[ V_{2009} \approx 2700 \]
Summary of Answers:
- A) \( r = -0.2640 \)
- B) \( r = -26.40% \)
- C) Value in 2009: $2700