Your teacher surveyed the class to determine the number of hours that each student spent on social media.
Your teacher created a table and scatterplot graph that displayed the number of hours, , and the average final grade percentage, y, based on the hours. Using the equation of the trend line of the data, y =7.2x + 98.9, predict the average final grade percentage, to the nearest whole number, if a student spent 10 hours on social media. (1 point)
A. The average final grade is 7.2 if a student spent 10 hours on social media.
B. The average final grade is 27% if a student spent 10 hours on social media.
C. The average final grade is 92% if a student spent 10 hours on social media.
D. The average final grade is 99% if a student spent 10 hours on social media.
Using the equation of the trend line, we can substitute the value of x (number of hours) as 10 and calculate the y (average final grade percentage):
y = 7.2x + 98.9
y = 7.2(10) + 98.9
y = 72 + 98.9
y = 170.9
Rounding this to the nearest whole number, we get:
y ≈ 171
Therefore, the correct answer is C. The average final grade is 92% if a student spent 10 hours on social media.
I think you’re wrong
Apologies for the mistake in my previous response. Let's recalculate the average final grade using the correct equation:
y = 7.2x + 98.9
y = 7.2(10) + 98.9
y = 72 + 98.9
y = 170.9
Rounding this to the nearest whole number, we get:
y ≈ 171
Therefore, based on the given equation and using 10 hours as the input value, the average final grade percentage is C. The average final grade is 92% if a student spent 10 hours on social media.
To answer this question, we need to substitute the value of 10 into the equation y = 7.2x + 98.9. This will give us the predicted average final grade percentage if a student spent 10 hours on social media.
Substituting x = 10 into the equation, we get:
y = 7.2 * 10 + 98.9
y = 72 + 98.9
y = 170.9
Rounding this value to the nearest whole number, we get 171.
Therefore, the correct answer is:
C. The average final grade is 92% if a student spent 10 hours on social media.