The following table shows students test scores on the first two tests in an introductory physics class.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline \multicolumn{10}{|c|}{ Physics Test Scores } \\
\hline \begin{tabular}{c}
First \\
test, $x$
\end{tabular} & 41 & 74 & 88 & 71 & 40 & 61 & 53 & 91 & 47 & 46 & 76 & 73 \\
\hline \begin{tabular}{c}
Second \\
test, $y$
\end{tabular} & 48 & 77 & 87 & 72 & 51 & 60 & 57 & 95 & 54 & 58 & 75 & 69 \\
\hline
\end{tabular}
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Step 2 of 2: If a student scored a 49 on his first test, make a prediction for his score on the second test. Assume the regression equation is appropriate for prediction. Round your answer to two decimal places, if necessary.
Answer
The final predicted score for the student on the second test is \(\boxed{55.38}\)
Step 1 ：To make a prediction for the student's score on the second test based on their score on the first test, we need to use the regression equation.
Step 2 ：However, the regression equation is not provided in the question. We need to calculate the regression equation using the given data for the first and second test scores.
Step 3 ：Using the given data, we calculate the predicted score for a student who scored 49 on the first test.
Step 4 ：The predicted score is calculated to be 55.37890005913661.
Step 5 ：We round the predicted score to two decimal places.
Step 6 ：The final predicted score for the student on the second test is \(\boxed{55.38}\)
