Let's walk through the solution to determine the best fit equation for the given data and then use it to make a prediction for the savings deposit at month 20.
The given data is:
[tex]\[
\begin{array}{|c|c|}
\hline \text{Month} & \text{Savings Deposit (S)} \\
\hline 0 & 10.80 \\
\hline 1 & 8.20 \\
\hline 2 & 6.00 \\
\hline 4 & 2.80 \\
\hline 7 & 1.00 \\
\hline 12 & 6.00 \\
\hline
\end{array}
\][/tex]
We have the following choices for the best fit equations and the approximate savings deposit at month 20:
A. [tex]\( y = 0.2 x^2 - 2.8 x + 10.8 \)[/tex] and approximately \[tex]$35
B. \( y = 0.7 x^2 - 4.9 x + 11.6 \) and approximately \$[/tex]35
C. [tex]\( y = 0.2 x^2 - 2.8 x + 10.8 \)[/tex] and approximately \[tex]$194
D. \( y = 0.7 x^2 - 4.9 x + 11.6 \) and approximately \$[/tex]194
To determine the best fit equation, we consider the given choices and calculate the savings deposit value at month 20 using each equation.
First, let's consider equation [tex]\( y = 0.2x^2 - 2.8x + 10.8 \)[/tex].
To find the savings deposit at month 20 ([tex]\(x = 20\)[/tex]):
[tex]\[
y = 0.2(20)^2 - 2.8(20) + 10.8
\][/tex]
From this equation, we know that [tex]\( y \approx 34.8 \)[/tex].
Next, let's check the second form of the equation, [tex]\( y = 0.7x^2 - 4.9x + 11.6 \)[/tex].
To find the savings deposit at month 20 ([tex]\(x = 20\)[/tex]):
[tex]\[
y = 0.7(20)^2 - 4.9(20) + 11.6
\][/tex]
After performing the calculations, we obtain a significantly different numerical result.
Given the options and the calculations, we notice that for the choice of equation [tex]$y = 0.2x^2 - 2.8x + 10.8$[/tex]:
- It returns approximately \[tex]$34.8 at month 20
- There is no fit consistent with the other approximations.
Therefore, the correct answer is:
\[
\boxed{A. \, y = 0.2 x^2 - 2.8 x + 10.8 \, \text{and approximately} \, \$[/tex]35 }
\]
