To solve this problem, we need to compute an exponential regression model in the form:

$y = a(b)^x$

Steps:

1. Determine $a$:

   • $a$ represents the initial value of $y$ when $x = 0$. From the table, $a = 18200$.

2. Determine $b$:

   • $b$ is the growth (or decay) factor and can be calculated as: $b = \sqrt[n]{\frac{y_n}{y_0}}$ Here, $y_0$ is the initial value (18200), and $y_n$ is the value at $x = n$. We will use several values to calculate $b$.

3. Calculate the correlation coefficient $r$:

   • This measures how well the exponential model fits the data.

Let me calculate these values for you.Here are the results for the exponential regression:

1. $a = 18200$ (initial value when $x = 0$).
2. $b = 0.88$ (decay factor, rounded to the nearest hundredth).
3. $r = -0.99$ (correlation coefficient, indicating a strong negative relationship).

Regression equation:

$y = 18200(0.88)^x$

Let me know if you need detailed explanations for any step or additional analysis!

Related Questions:

1. How does the correlation coefficient $r$ affect the interpretation of the model's fit?
2. Can you calculate the expected value of the car after 5 years using the regression equation?
3. How would the regression change if the decay factor $b$ were larger than 1?
4. Why is exponential regression appropriate for this data instead of linear regression?
5. How can we verify the model's accuracy with residuals?

Tip:

In exponential models, the decay factor $b$ gives a direct measure of the rate of decrease (or increase) per unit of $x$.