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To solve this problem, we need to follow these steps:

### 1. **Determine the Linear Regression Equation**
A linear regression equation has the form:
\[
V(n) = mn + b
\]
Where:
- $n$ is the number of years after 1988,
- $V(n)$ is the value of the investment at year $n$,
- $m$ is the slope (rate of change of the investment per year),
- $b$ is the y-intercept (value of the investment at $n = 0$, i.e., in 1988).

### 2. **Find the Slope $m$**
The slope $m$ is calculated using two points $(n_1, V(n_1))$ and $(n_2, V(n_2))$:
\[
m = \frac{V(n_2) - V(n_1)}{n_2 - n_1}
\]
We can choose any two points from the table. Let's use the first and the last data points:
- For $n_1 = 1$, $V(1) = 10181$,
- For $n_2 = 19$, $V(19) = 5927.05$.

Now, calculate the slope $m$:
\[
m = \frac{5927.05 - 10181}{19 - 1}
\]
\[
m = \frac{-4253.95}{18}
\]
\[
m \approx -236.33
\]

### 3. **Find the y-intercept $b$**
Now that we have the slope $m$, we can substitute one point into the linear equation to solve for $b$. Let's use the point $(n = 1, V(n) = 10181)$:
\[
10181 = -236.33(1) + b
\]
\[
10181 = -236.33 + b
\]
\[
b = 10181 + 236.33 = 10417.33
\]

### 4. **Write the Linear Regression Equation**
Now we have both $m$ and $b$, so the equation becomes:
\[
V(n) = -236.33n + 10417.33
\]

### 5. **Answer the Questions**
- **The value of the investment in the year 1988** is given by $b$, which is the value at $n = 0$:
\[
V(0) = 10417.33
\]

- **The rate of change of the investment** is the slope $m$, which is $-236.33$ dollars per year. The value is decreasing by this amount each year.

#### Final answers:
- The value of the investment in 1988: $\$10,417.33$.
- The value of the investment is **decreasing** at a rate of $\$236.33$ per year.

Would you like more details on any step or further clarification?

Here are 5 questions that could extend this topic:
1. How can we use this linear model to predict future values of the investment?
2. What does the negative slope imply about the trend in investment value over time?
3. Could a different type of regression (e.g., exponential) model this data more accurately?
4. What happens to the investment value after 19 years, based on this model?
5. How would we find the point at which the investment becomes zero?

**Tip:** When using linear regression, the slope tells you how much the dependent variable changes for every unit increase in the independent variable.

Determine the linear regression equation that models the set of data, then use this equation to find the value of the investment in 1988 and the rate of decrease per year.

Learn how to calculate a linear regression equation for investment data starting from 1988. This problem involves finding the slope and intercept, and determining how the value of the investment changes over time, with key data points including the value of $10181 at year 1 and $5927.05 at year 19.

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