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Use the equation for the line of best fit to predict the salary for an employee with 7 years of experience.

\begin{tabular}{|l|l|}
\hline
Years of experience & Salary \\
\hline
0 & [tex]$\$[/tex] 50,000[tex]$ \\
\hline
1 & $[/tex]\[tex]$ 52,150$[/tex] \\
\hline
2 & [tex]$\$[/tex] 54,260[tex]$ \\
\hline
3 & $[/tex]\[tex]$ 56,285$[/tex] \\
\hline
4 & [tex]$\$[/tex] 58,820[tex]$ \\
\hline
5 & $[/tex]\[tex]$ 60,126$[/tex] \\
\hline
\end{tabular}

A. [tex]$\$[/tex] 68,150[tex]$
B. $[/tex]\[tex]$ 64,616$[/tex]
C. [tex]$\$[/tex] 62,738[tex]$
D. $[/tex]\[tex]$ 62,540$[/tex]

Sagot :

To predict the salary for an employee with 7 years of experience, we need to determine the equation of the line of best fit for the given data points. The line of best fit can be represented by the linear equation:

[tex]\[ \text{Salary} = \text{slope} \times (\text{Years of experience}) + \text{intercept} \][/tex]

Given the provided data points:

[tex]\[ \begin{array}{|c|c|} \hline \text{Years of experience} & \text{Salary} \\ \hline 0 & \$50,000 \\ 1 & \$52,150 \\ 2 & \$54,260 \\ 3 & \$56,285 \\ 4 & \$58,820 \\ 5 & \$60,126 \\ \hline \end{array} \][/tex]

we first calculate the slope and intercept of the line. The determined slope and intercept are:

- The slope is approximately [tex]\( 2076.142857142858 \)[/tex]
- The intercept is approximately [tex]\( 50083.14285714284 \)[/tex]

Next, we substitute these values into the linear equation and use [tex]\( 7 \)[/tex] years of experience to find the predicted salary:

[tex]\[ \text{Predicted Salary} = 2076.142857142858 \times 7 + 50083.14285714284 \][/tex]

Calculating this we get:

[tex]\[ \text{Predicted Salary} \approx 2076.142857142858 \times 7 + 50083.14285714284 \][/tex]
[tex]\[ \text{Predicted Salary} \approx 14532.999999999906 + 50083.14285714284 \][/tex]
[tex]\[ \text{Predicted Salary} \approx 64616.14285714284 \][/tex]

Therefore, the predicted salary for an employee with 7 years of experience is:

[tex]\[\$64,616\][/tex]

So the correct answer is:

[tex]\[\$64,616\][/tex]
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