Problem
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} Copy Data 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.
Solution
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 \(\boxed{55.38}\)
