Questions: A car was valued at 28,000 in the year 1994. The value depreciated to 13,000 by the year 2005. A) What was the annual rate of change between 1994 and 2005? r= 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 2009? value = Round to the nearest 50 dollars.

Solution Steps

A) To find the annual rate of change, we can use the formula for exponential decay: \( V = V_0 \times (1 - r)^t \), where \( V \) is the final value, \( V_0 \) is the initial value, \( r \) is the rate of change, and \( t \) is the time in years. We can solve for \( r \) using the given values. B) Convert the rate of change \( r \) to a percentage by multiplying by 100. C) Use the rate of change found in part A to predict the value of the car in 2009 using the same exponential decay formula.

To find the annual rate of change \( r \) between 1994 and 2005, we use the formula for exponential decay:

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

Substituting the known values:

\[ 13000 = 28000 \times (1 - r)^{11} \]

Solving for \( r \), we find:

\[ 1 - r = \left( \frac{13000}{28000} \right)^{\frac{1}{11}} \implies r \approx 0.0674 \]

To express the annual rate of change as a percentage, we multiply \( r \) by 100:

\[ r_{\text{percentage}} = 0.0674 \times 100 \approx 6.74\% \]

Using the same rate of change, we can predict the value of the car in 2009. The time from 1994 to 2009 is 15 years. Thus, we calculate:

\[ V_{2009} = 28000 \times (1 - 0.0674)^{15} \approx 9830.81 \]

Rounding to the nearest 50 dollars gives:

\[ V_{2009, \text{rounded}} \approx 9850 \]

Final Answer