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 $V = an + b$, where $a$ is the slope and $b$ is the y - intercept. We can use a statistical software or a calculator with linear - regression functionality. Let $x_i$ be the values of $n$ and $y_i$ be the values of $V$. The formulas for $a$ and $b$ are: [a=\frac{n\sum_{i = 1}^{k}x_iy_i-\sum_{i = 1}^{k}x_i\sum_{i = 1}^{k}y_i}{n\sum_{i = 1}^{k}x_i^{2}-(\sum_{i = 1}^{k}x_i)^{2}}] [b=\frac{\sum_{i = 1}^{k}y_i - a\sum_{i = 1}^{k}x_i}{n}] where $n = 6$ (the number of data points), $x_1 = 1,x_2 = 3,x_3 = 7,x_4 = 12,x_5 = 14,x_6 = 19$ and $y_1 = 19607,y_2 = 19816.58,y_3 = 21449,y_4 = 23673.52,y_5 = 23598,y_6 = 23876.35$. First, calculate the necessary sums: (\sum_{i = 1}^{6}x_i=1 + 3+7 + 12+14+19=56) (\sum_{i = 1}^{6}y_i=19607+19816.58+21449+23673.52+23598+23876.35 = 121020.45) (\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\times19607+3\times19816.58+7\times21449+12\times23673.52+14\times23598+19\times23876.35) [=19607+59449.74+150143+284082.24+330372+453650.65=1307304.63]

Step2: Calculate the slope $a$

[a=\frac{6\times1307304.63-56\times121020.45}{6\times760 - 56^{2}}] [=\frac{7843827.78-6777145.2}{4560 - 3136}] [=\frac{1066682.58}{1424}\approx749.07]

Step3: Calculate the y - intercept $b$

[b=\frac{121020.45-749.07\times56}{6}] [=\frac{121020.45 - 41947.92}{6}] [=\frac{79072.53}{6}\approx13178.76] So the linear - regression equation is $V = 749.07n+13178.76$. In 1985, $n = 0$. Substitute $n = 0$ into the equation: [V=749.07\times0 + 13178.76=13178.76] The rate of change is the slope of the linear - regression line, which is $a = 749.07$.

Answer:

The value of the investment in 1985 was $$13178.76$. The value of the investment is increasing at a rate of $$749.07$ per year.