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\begin{tabular}{|l|l|}
\hline
\begin{tabular}{l}
Speed \\
(mph)
\end{tabular} & \begin{tabular}{l}
Stopping \\
distance \\
(ft)
\end{tabular} \\
\hline
10 & 12.5 \\
\hline
20 & 36.0 \\
\hline
30 & 69.5 \\
\hline
40 & 114.0 \\
\hline
50 & 169.5 \\
\hline
60 & 249.0 \\
\hline
70 & 325.5 \\
\hline
\end{tabular}

Use the quadratic regression feature of a graphing calculator to find a quadratic model. Round to the nearest hundredths place.

A. [tex]y=0.06 x^2+0.31 x+4[/tex]
B. [tex]y=-4.03 x^2+0.32 x+8.19[/tex]
C. [tex]y=0.06 x^2-0.31 x-4[/tex]
D. [tex]y=4.03 x^2-0.32 x-8.19[/tex]


Sagot :

To find a quadratic model for the given data points using quadratic regression, follow these steps:

1. List the Data Points:
- Speed (x): 10, 20, 30, 40, 50, 60, 70
- Stopping Distance (y): 12.5, 36.0, 69.5, 114.0, 169.5, 249.0, 325.5

2. Formulate a Quadratic Model:
- The general form of a quadratic model is [tex]\( y = ax^2 + bx + c \)[/tex].

3. Determine the Coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
- Using quadratic regression methods, we can determine the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex].
- After applying quadratic regression to the given data points and rounding the coefficients to the nearest hundredths place, we get:
- [tex]\( a = 0.06 \)[/tex]
- [tex]\( b = 0.31 \)[/tex]
- [tex]\( c = 4.0 \)[/tex]

4. Construct the Model Equation:
- Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the quadratic model [tex]\( y = ax^2 + bx + c \)[/tex]:
[tex]\[ y = 0.06x^2 + 0.31x + 4.0 \][/tex]

5. Compare with Given Options:
- Option a: [tex]\( y = 0.06x^2 + 0.31x + 4 \)[/tex]
- Option b: [tex]\( y = -4.03x^2 + 0.32x + 8.19 \)[/tex]
- Option c: [tex]\( y = 0.06x^2 - 0.31x - 4 \)[/tex]
- Option d: [tex]\( y = 4.03x^2 - 0.32x - 8.19 \)[/tex]

6. Identify Correct Option:
- The model we constructed, [tex]\( y = 0.06x^2 + 0.31x + 4.0 \)[/tex], matches Option a.

Therefore, the correct answer is:
a. [tex]\( y = 0.06x^2 + 0.31x + 4 \)[/tex].
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