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To solve this problem, we need to determine the linear regression equation that models the data and answers the questions based on that regression.

The table shows the following values:

| $ n $ (years after 1988) | 3 | 7 | 12 | 14 | 19 |
|---------------------------|----|----|-----|-----|-----|
| $ V(n) $ (investment value in dollars) | 10181 | 9548.14 | 8867 | 8005.16 | 7334 | 5927.05 |

The form of a linear regression equation is:

\[
V(n) = mn + b
\]

Where:
- $ m $ is the slope (rate of decrease of the investment per year),
- $ b $ is the y-intercept (value of the investment in 1988, or when $ n = 0 $).

### Step 1: Find the slope $ m $
The slope is calculated using two data points $(n_1, V(n_1))$ and $(n_2, V(n_2))$ from the table. Let's choose the first and last data points:

\[
m = \frac{V(n_2) - V(n_1)}{n_2 - n_1} = \frac{5927.05 - 10181}{19 - 3}
\]

### Step 2: Solve for the y-intercept $ b $
Once we have the slope, we can use one of the points to solve for $ b $ by substituting it into the equation $ V(n) = mn + b $.

Let me calculate the slope and equation.The slope $ m $ of the linear regression equation is approximately $ -265.87 $. This means the investment value decreases by about $265.87 per year.

### Step 3: Find the y-intercept $ b $
Now, we will substitute one of the points (let's use $ n = 3 $, $ V(3) = 10181 $) into the equation $ V(n) = mn + b $ to solve for $ b $:

\[
10181 = -265.87(3) + b
\]

Let me calculate the y-intercept $ b $.The y-intercept $ b $ is approximately $ 10978.62 $. This represents the estimated value of the investment in the year 1988 (when $ n = 0 $).

### The linear regression equation is:

\[
V(n) = -265.87n + 10978.62
\]

### Step 4: Answer the questions
1. **The value of the investment in 1988** is approximately **$10,978.62**.
2. **The rate at which the value decreases** is **$265.87 per year**.

Would you like to see more details, or do you have any other questions?

Here are 5 relative questions to expand on this information:
1. What is the projected value of the investment in 10 years after 1988?
2. How can we use this regression model to predict future values beyond the data points given?
3. What happens to the investment value after 30 years, according to this model?
4. How would the equation change if the investment's value increased instead of decreased?
5. How accurate are linear regression models for predicting long-term trends in investments?

**Tip:** Always ensure to check if a linear regression model is the best fit by analyzing the trend in data.

Determine the linear regression equation that models the set of data above, and use this equation to answer the questions. The table shows the value, V, of an investment (in dollars) n years after 1988 for various values of n.

Learn how to determine the linear regression equation that models the investment values after 1988, using key algebraic concepts like slope and intercept. The table shows values for years 3, 7, 12, 14, and 19 after 1988.

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