Question

question 6
1.4/7 pts 3 98 details
linear regression
the table below shows the value, v, of an investment (in dollars) n years after 1995.

n	1	3	7	12	14	19
v	6148	6511.12	7636	9142.28	9372	10081.4

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 tenth as needed.
based on this regression model, the value of this investment was $ in the year 1995.
based on the regression model, the value of this investment is select an answer at a rate of $ per year.
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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 calculator or software with linear - regression capabilities. Using a calculator (e.g., TI - 84 Plus), input the data points $(n_1,V_1),(n_2,V_2),\cdots,(n_6,V_6)$ where $n_1 = 1,V_1=6148,n_2 = 3,V_2 = 6511.12,\cdots,n_6=19,V_6 = 10081.4$.

Step2: Find the coefficients

After running the linear - regression function on the calculator, we get $a\approx249.9$ and $b\approx5898.1$. So the linear - regression equation is $V=249.9n + 5898.1$.

Step3: Find the value in 1995

In 1995, $n = 0$. Substitute $n = 0$ into the equation $V=249.9n + 5898.1$. Then $V=249.9\times0+5898.1 = 5898.1\approx5898.1$.

Step4: Find the rate of change

The rate of change of the investment value per year is the slope of the linear - regression line. The slope $a = 249.9\approx249.9$.

Answer:

The value of the investment in 1995 was $\$5898.1$. The value of the investment is increasing at a rate of $\$249.9$ per year.