Let's break down the problem in steps:

1. Identify the Data: The table provides retail prices for a car from 1998 to 2002, with x = 1 representing the year 1998, and x increments by 1 for each subsequent year.

   • Data points:
     • $(x = 1, y = 15375)$
     • $(x = 2, y = 11554)$
     • $(x = 3, y = 10567)$
     • $(x = 4, y = 9916)$
     • $(x = 5, y = 8883)$

2. Find the Regression Equation: To find the regression equation $y = mx + b$ (a linear fit), we would calculate:

   • Slope (m) and intercept (b) using the least squares method, based on the points above.

3. Predict for 2004: Since 1998 is represented by $x = 1$, the year 2004 corresponds to $x = 7$.

4. Determine Year for Value of $2486: Set $y = 2486$ in the regression equation and solve for $x$.

I'll calculate the regression equation now.The regression equation, based on the data, is:

$y = -1462.2x + 15645.6$

Using this equation:

1. Prediction for 2004: In 2004 ($x = 7$), the average retail price of the car is estimated to be approximately $5410.20.

2. Year when the car's value will reach $2486: Solving for $x$ when $y = 2486$ gives an approximate value of $x \approx 9$. This corresponds to the year 2006.

Would you like further details on these calculations, or do you have other questions?

Related Questions:

1. How do we calculate the slope and intercept of a regression line?
2. Why does the linear model assume a constant rate of depreciation?
3. What other methods could predict prices if the data isn't linear?
4. How does the correlation coefficient ($r$) relate to the accuracy of predictions?
5. What happens if we need to predict prices far outside the given data range?

Tip:

When using a regression equation to predict values outside the given data range, be cautious of "extrapolation," as the model may not hold accurately far beyond the observed data points.