Answer :
To determine which rational function best models the data provided in the table, follow these steps:
1. Extract the Given Data Points:
We have four data points for time [tex]\(x\)[/tex] and corresponding average speed [tex]\(y\)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time, } x \text{ (hours)} & \text{Average Speed, } y \text{ (miles per hour)} \\ \hline 12 & 8 \\ \hline 16 & 6 \\ \hline 10 \frac{2}{3} & 9 \\ \hline 18 & 5 \frac{1}{3} \\ \hline \end{array} \][/tex]
Converting the mixed fractions into decimals:
[tex]\[ 10 \frac{2}{3} = 10.67 \quad \text{and} \quad 5 \frac{1}{3} = 5.33 \][/tex]
2. Consider the Candidate Rational Functions:
The two rational functions given are:
[tex]\[ f_1(x) = \frac{x}{96} \][/tex]
[tex]\[ f_2(x) = \frac{2x}{3} \][/tex]
3. Calculate the Predicted Speed Values Using the Functions:
For [tex]\( f_1(x) = \frac{x}{96} \)[/tex]:
[tex]\[ f_1(12) = \frac{12}{96} = 0.125 \][/tex]
[tex]\[ f_1(16) = \frac{16}{96} = 0.1667 \][/tex]
[tex]\[ f_1(10.67) = \frac{10.67}{96} = 0.1111 \][/tex]
[tex]\[ f_1(18) = \frac{18}{96} = 0.1875 \][/tex]
For [tex]\( f_2(x) = \frac{2x}{3} \)[/tex]:
[tex]\[ f_2(12) = \frac{2 \cdot 12}{3} = 8 \][/tex]
[tex]\[ f_2(16) = \frac{2 \cdot 16}{3} = 10.67 \][/tex]
[tex]\[ f_2(10.67) = \frac{2 \cdot 10.67}{3} = 7.11 \][/tex]
[tex]\[ f_2(18) = \frac{2 \cdot 18}{3} = 12 \][/tex]
4. Calculate the Mean Squared Error (MSE) for Each Function:
The MSE is calculated as follows:
[tex]\[ \text{MSE} = \frac{1}{n} \sum_{i=1}^{n} (y_i - \hat{y}_i)^2 \][/tex]
For [tex]\(f_1(x)\)[/tex]:
[tex]\[ \text{MSE}_{f_1} = \frac{1}{4} \left[(8 - 0.125)^2 + (6 - 0.1667)^2 + (9 - 0.1111)^2 + (5.33 - 0.1875)^2\right] \approx 50.38 \][/tex]
For [tex]\(f_2(x)\)[/tex]:
[tex]\[ \text{MSE}_{f_2} = \frac{1}{4} \left[(8 - 8)^2 + (6 - 10.67)^2 + (9 - 7.11)^2 + (5.33 - 12)^2\right] \approx 17.45 \][/tex]
5. Compare the MSE Values:
[tex]\[ \text{MSE}_{f_1} \approx 50.38 \][/tex]
[tex]\[ \text{MSE}_{f_2} \approx 17.45 \][/tex]
Since [tex]\( \text{MSE}_{f_2} \)[/tex] is less than [tex]\( \text{MSE}_{f_1} \)[/tex], the function [tex]\( f_2(x) = \frac{2x}{3} \)[/tex] has a lower mean squared error and thus, better fits the data.
Hence, the rational function [tex]\( y = \frac{2x}{3} \)[/tex] best models the data in the table.
1. Extract the Given Data Points:
We have four data points for time [tex]\(x\)[/tex] and corresponding average speed [tex]\(y\)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time, } x \text{ (hours)} & \text{Average Speed, } y \text{ (miles per hour)} \\ \hline 12 & 8 \\ \hline 16 & 6 \\ \hline 10 \frac{2}{3} & 9 \\ \hline 18 & 5 \frac{1}{3} \\ \hline \end{array} \][/tex]
Converting the mixed fractions into decimals:
[tex]\[ 10 \frac{2}{3} = 10.67 \quad \text{and} \quad 5 \frac{1}{3} = 5.33 \][/tex]
2. Consider the Candidate Rational Functions:
The two rational functions given are:
[tex]\[ f_1(x) = \frac{x}{96} \][/tex]
[tex]\[ f_2(x) = \frac{2x}{3} \][/tex]
3. Calculate the Predicted Speed Values Using the Functions:
For [tex]\( f_1(x) = \frac{x}{96} \)[/tex]:
[tex]\[ f_1(12) = \frac{12}{96} = 0.125 \][/tex]
[tex]\[ f_1(16) = \frac{16}{96} = 0.1667 \][/tex]
[tex]\[ f_1(10.67) = \frac{10.67}{96} = 0.1111 \][/tex]
[tex]\[ f_1(18) = \frac{18}{96} = 0.1875 \][/tex]
For [tex]\( f_2(x) = \frac{2x}{3} \)[/tex]:
[tex]\[ f_2(12) = \frac{2 \cdot 12}{3} = 8 \][/tex]
[tex]\[ f_2(16) = \frac{2 \cdot 16}{3} = 10.67 \][/tex]
[tex]\[ f_2(10.67) = \frac{2 \cdot 10.67}{3} = 7.11 \][/tex]
[tex]\[ f_2(18) = \frac{2 \cdot 18}{3} = 12 \][/tex]
4. Calculate the Mean Squared Error (MSE) for Each Function:
The MSE is calculated as follows:
[tex]\[ \text{MSE} = \frac{1}{n} \sum_{i=1}^{n} (y_i - \hat{y}_i)^2 \][/tex]
For [tex]\(f_1(x)\)[/tex]:
[tex]\[ \text{MSE}_{f_1} = \frac{1}{4} \left[(8 - 0.125)^2 + (6 - 0.1667)^2 + (9 - 0.1111)^2 + (5.33 - 0.1875)^2\right] \approx 50.38 \][/tex]
For [tex]\(f_2(x)\)[/tex]:
[tex]\[ \text{MSE}_{f_2} = \frac{1}{4} \left[(8 - 8)^2 + (6 - 10.67)^2 + (9 - 7.11)^2 + (5.33 - 12)^2\right] \approx 17.45 \][/tex]
5. Compare the MSE Values:
[tex]\[ \text{MSE}_{f_1} \approx 50.38 \][/tex]
[tex]\[ \text{MSE}_{f_2} \approx 17.45 \][/tex]
Since [tex]\( \text{MSE}_{f_2} \)[/tex] is less than [tex]\( \text{MSE}_{f_1} \)[/tex], the function [tex]\( f_2(x) = \frac{2x}{3} \)[/tex] has a lower mean squared error and thus, better fits the data.
Hence, the rational function [tex]\( y = \frac{2x}{3} \)[/tex] best models the data in the table.