The table below shows the value, V, of an investment (in dollars) n years after 1991.



**n**

1

3

7

12

14

19

**V(n)**

14393

13503.42

12551

11346.48

10402

8423.65

Determine the linear regression equation that models the set of data above, and use this equation to answer the questions below. Round to the nearest hundredth as needed.

Based on this regression model, the value of this investment was $ in the year 1991.

Based on the regression model, the value of this investment is Select an answer decreasing increasing 
at a rate of $ per year.

To find the linear regression equation that models the set of data, we need to calculate the best-fit line in the form:

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

where:
- $ V(n) $ is the value of the investment,
- $ n $ is the number of years after 1991,
- $ m $ is the slope of the line (which shows the rate of change),
- $ b $ is the y-intercept (which represents the value of the investment in 1991, when $ n = 0 $).

### Step 1: Gather Data Points

The data points from the table are as follows (with $ n $ as the number of years since 1991):

\[
(n, V(n)) = \{(1, 14393), (3, 13503.42), (7, 12551), (12, 11346.48), (14, 10402), (19, 8423.65)\}
\]

### Step 2: Calculate the Linear Regression Equation

To compute the linear regression equation, we need to calculate the slope $ m $ and the y-intercept $ b $. These can be computed using formulas for the slope and intercept of a line based on the least-squares method:

\[
m = \frac{n \sum{(n_i V_i)} - (\sum n_i)(\sum V_i)}{n \sum n_i^2 - (\sum n_i)^2}
\]

\[
b = \frac{\sum V_i - m \sum n_i}{n}
\]

To simplify, I'll perform the calculations to find the slope and intercept for you.

### Step 3: Interpret Results

- **The y-intercept $ b $** will give the value of the investment in 1991.
- **The slope $ m $** will tell us whether the investment is increasing or decreasing and at what rate per year.

Let me calculate this for you.The linear regression equation that models the data is:

\[
V(n) = -313.99n + 14700.50
\]

### Interpretation:
1. **The value of the investment in 1991 (when $ n = 0 $)**:
   - The y-intercept is $ b = 14700.50 $, so the value of the investment in 1991 was approximately **$14,700.50**.
   
2. **Rate of change**:
   - The slope $ m = -313.99 $ indicates that the investment is **decreasing** at a rate of approximately **$313.99 per year**.

Would you like further details or have any questions?

### Related Questions:
1. How can the model be used to predict the investment value in future years?
2. Can we determine when the investment might reach zero based on this model?
3. How reliable is the linear model for long-term predictions?
4. Is there any way to check the accuracy of the regression model?
5. Could this model be extended to cover nonlinear trends in the data?

### Tip:
Always check the context when using linear regression, as real-world investments might not follow a strictly linear pattern over long periods.

The table below shows the value, V, of an investment (in dollars) n years after 1991.

n: 1, 3, 7, 12, 14, 19
V(n): 14393, 13503.42, 12551, 11346.48, 10402, 8423.65

Determine the linear regression equation that models the set of data above, and use this equation to answer the questions below. Round to the nearest hundredth as needed.

Based on this regression model, the value of this investment was $___ in the year 1991.

Based on the regression model, the value of this investment is ___ (decreasing/increasing) at a rate of $___ per year.

Learn how to calculate a linear regression model for an investment's value over time. Using a set of data points, we determine the slope and y-intercept, giving us a formula that shows the value of the investment in 1991 and the rate of decrease over the years. Suitable for students in Grades 11-12.

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