Answer :
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: \[ \boxed{A. \, y = 0.2 x^2 - 2.8 x + 10.8 \, \text{and approximately} \, \$[/tex]35 }
\]
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: \[ \boxed{A. \, y = 0.2 x^2 - 2.8 x + 10.8 \, \text{and approximately} \, \$[/tex]35 }
\]