linear regression\nthe table below shows the value, v, of an investment (in dollars) n years after…

Explanation:

Step1: Recall linear - regression formula

The linear - regression equation is of the form $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. For our data, let $n$ be the independent variable (years after 1989) and $V(n)$ be the dependent variable (value of the investment). We can use a statistical software or a calculator with linear - regression capabilities. Using the formula for the slope $m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}$ and $b=\bar{y}-m\bar{x}$, where $x_i$ are the values of $n$ and $y_i$ are the values of $V(n)$. Let $x=[1,3,7,12,14,19]$ and $y = [16588,15644.72,14716,13550.68,12532,10423.4]$. $n = 6$ $\sum_{i=1}^{6}x_i=1 + 3+7+12+14+19=56$ $\sum_{i=1}^{6}y_i=16588 + 15644.72+14716+13550.68+12532+10423.4 = 83454.8$ $\sum_{i=1}^{6}x_i^{2}=1^{2}+3^{2}+7^{2}+12^{2}+14^{2}+19^{2}=1 + 9+49+144+196+361 = 760$ $\sum_{i=1}^{6}x_iy_i=1\times16588+3\times15644.72+7\times14716+12\times13550.68+14\times12532+19\times10423.4$ $=16588+46934.16+103012+162608.16+175448+198044.6=692634.92$

$m=\frac{6\times692634.92-56\times83454.8}{6\times760 - 56^{2}}$ $=\frac{4155809.52-4673468.8}{4560 - 3136}$ $=\frac{- 517659.28}{1424}\approx - 363.40$

$\bar{x}=\frac{56}{6}\approx9.33$ $\bar{y}=\frac{83454.8}{6}\approx13909.13$

$b=\bar{y}-m\bar{x}=13909.13-(-363.40)\times9.33=13909.13 + 3390.52=17299.65$

So the linear - regression equation is $V(n)=-363.40n + 17299.65$

Step2: Find the value in 1989

In 1989, $n = 0$. Substitute $n = 0$ into the equation $V(n)=-363.40n + 17299.65$. $V(0)=-363.40\times0+17299.65 = 17299.65$

Step3: Determine the rate of change

The slope of the linear - regression equation represents the rate of change. The slope $m=-363.40$. The value of the investment is decreasing at a rate of $$363.40$ per year.

Answer:

17299.65 Decreasing 363.40