Answer :
Let's break down the problem step-by-step to determine which equation best models the given data.
Given the data:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline y & 32 & 67 & 79 & 91 & 98 & 106 & 114 & 120 & 126 & 132 \\ \hline \end{array} \][/tex]
### Step 1: Calculate the differences between consecutive [tex]\( y \)[/tex] values.
Let's look at the differences [tex]\(\Delta y\)[/tex] between consecutive [tex]\( y \)[/tex] values:
[tex]\[ \begin{array}{|c|c|} \hline \Delta y & \\ \hline 67 - 32 & = 35 \\ \hline 79 - 67 & = 12 \\ \hline 91 - 79 & = 12 \\ \hline 98 - 91 & = 7 \\ \hline 106 - 98 & = 8 \\ \hline 114 - 106 & = 8 \\ \hline 120 - 114 & = 6 \\ \hline 126 - 120 & = 6 \\ \hline 132 - 126 & = 6 \\ \hline \end{array} \][/tex]
These differences are [tex]\([35, 12, 12, 7, 8, 8, 6, 6, 6]\)[/tex].
### Step 2: Calculate the mean difference.
To find the mean difference, sum the differences and divide by the number of differences.
[tex]\[ \text{Mean difference} = \frac{35 + 12 + 12 + 7 + 8 + 8 + 6 + 6 + 6}{9} = \approx 11.11 \][/tex]
### Step 3: Check the potential model [tex]\( y = 33x + 32.7 \)[/tex].
Calculate [tex]\( y \)[/tex] values using [tex]\( y = 33x + 32.7 \)[/tex] for each given [tex]\( x \)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 33 \cdot 0 + 32.7 = 32.7 \\ \hline 1 & 33 \cdot 1 + 32.7 = 65.7 \\ \hline 2 & 33 \cdot 2 + 32.7 = 98.7 \\ \hline 3 & 33 \cdot 3 + 32.7 = 131.7 \\ \hline 4 & 33 \cdot 4 + 32.7 = 164.7 \\ \hline 5 & 33 \cdot 5 + 32.7 = 197.7 \\ \hline 6 & 33 \cdot 6 + 32.7 = 230.7 \\ \hline 7 & 33 \cdot 7 + 32.7 = 263.7 \\ \hline 8 & 33 \cdot 8 + 32.7 = 296.7 \\ \hline 9 & 33 \cdot 9 + 32.7 = 329.7 \\ \hline \end{array} \][/tex]
The calculated values are [tex]\([32.7, 65.7, 98.7, 131.7, 164.7, 197.7, 230.7, 263.7, 296.7, 329.7]\)[/tex].
### Conclusion
The equation that best models the given data points appears to be:
[tex]\[ y = 33x + 32.7 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{B. \, y = 33x + 32.7} \][/tex]
Given the data:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline y & 32 & 67 & 79 & 91 & 98 & 106 & 114 & 120 & 126 & 132 \\ \hline \end{array} \][/tex]
### Step 1: Calculate the differences between consecutive [tex]\( y \)[/tex] values.
Let's look at the differences [tex]\(\Delta y\)[/tex] between consecutive [tex]\( y \)[/tex] values:
[tex]\[ \begin{array}{|c|c|} \hline \Delta y & \\ \hline 67 - 32 & = 35 \\ \hline 79 - 67 & = 12 \\ \hline 91 - 79 & = 12 \\ \hline 98 - 91 & = 7 \\ \hline 106 - 98 & = 8 \\ \hline 114 - 106 & = 8 \\ \hline 120 - 114 & = 6 \\ \hline 126 - 120 & = 6 \\ \hline 132 - 126 & = 6 \\ \hline \end{array} \][/tex]
These differences are [tex]\([35, 12, 12, 7, 8, 8, 6, 6, 6]\)[/tex].
### Step 2: Calculate the mean difference.
To find the mean difference, sum the differences and divide by the number of differences.
[tex]\[ \text{Mean difference} = \frac{35 + 12 + 12 + 7 + 8 + 8 + 6 + 6 + 6}{9} = \approx 11.11 \][/tex]
### Step 3: Check the potential model [tex]\( y = 33x + 32.7 \)[/tex].
Calculate [tex]\( y \)[/tex] values using [tex]\( y = 33x + 32.7 \)[/tex] for each given [tex]\( x \)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 33 \cdot 0 + 32.7 = 32.7 \\ \hline 1 & 33 \cdot 1 + 32.7 = 65.7 \\ \hline 2 & 33 \cdot 2 + 32.7 = 98.7 \\ \hline 3 & 33 \cdot 3 + 32.7 = 131.7 \\ \hline 4 & 33 \cdot 4 + 32.7 = 164.7 \\ \hline 5 & 33 \cdot 5 + 32.7 = 197.7 \\ \hline 6 & 33 \cdot 6 + 32.7 = 230.7 \\ \hline 7 & 33 \cdot 7 + 32.7 = 263.7 \\ \hline 8 & 33 \cdot 8 + 32.7 = 296.7 \\ \hline 9 & 33 \cdot 9 + 32.7 = 329.7 \\ \hline \end{array} \][/tex]
The calculated values are [tex]\([32.7, 65.7, 98.7, 131.7, 164.7, 197.7, 230.7, 263.7, 296.7, 329.7]\)[/tex].
### Conclusion
The equation that best models the given data points appears to be:
[tex]\[ y = 33x + 32.7 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{B. \, y = 33x + 32.7} \][/tex]