Question

question 9 (1 point)
create the equation given the situation and the data below.

the paired data below consist of the test scores of 6 randomly selected students and the number of hours they studied for the test. use a graphing calculator to model the data with a linear function that predicts a student’s score as a function of the number of hours he or she studied.

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$$\begin{tabular}{c|ccccc} hours & 5 & 10 & 4 & 6 & 10 & 9 \\\\ score & 64 & 86 & 69 & 86 & 59 & 87 \\\\ \\end{tabular}$$

\bigcirc a) $y = 33.7 + 2.14x$

\bigcirc b) $y = 33.7 - 2.14x$

\bigcirc c) $y = 67.3 + 1.07x$

\bigcirc d) $y = -67.3 + 1.07x$

Explanation:

Step1: Analyze the trend

As hours studied (x) increase, scores (y) generally increase (e.g., 4 hours: 69, 5 hours: 64? Wait, no, 4→5: 69→64 (decrease), 5→6: 64→86 (increase), 6→9: 86→87 (increase), 9→10: 87→59 (decrease), 10→10: 59→86? Wait, data: Hours [5,10,4,6,10,9], Scores [64,86,69,86,59,87]. Let's calculate mean of x and y.
Mean of x: $\frac{5 + 10 + 4 + 6 + 10 + 9}{6} = \frac{44}{6} \approx 7.33$
Mean of y: $\frac{64 + 86 + 69 + 86 + 59 + 87}{6} = \frac{451}{6} \approx 75.17$

Step2: Check slope sign

Mostly, more hours should relate to higher scores (even with outliers). So slope should be positive. Eliminate B (negative slope) and D (negative intercept with positive slope, but mean y is ~75, mean x ~7.33: -67.3 +1.07*7.33 ≈ -67.3 +7.84 ≈ -59.46 ≠75.17). Now check A and C.

Step3: Test A: y=33.7+2.14x at x=7.33: 33.7 +2.14*7.33 ≈33.7 +15.69 ≈49.39 ≠75.17.

Test C: y=67.3+1.07x at x=7.33: 67.3 +1.07*7.33 ≈67.3 +7.84 ≈75.14, close to mean y. Also, check a point: x=10, y=67.3+10.7=78, but scores at x=10 are 86 and 59 (outlier 59). The linear model smooths. The correct model (from linear regression) should have positive slope, and intercept making mean y close. So C is better.

Answer:

C) \( y = 67.3 + 1.07x \)