b) Looking at the graph, the scores of quiz 2 are on the y axis while the scores of quiz 1 are on the y axis. Each samll box on both axes is 2 units. This means that half of a samll box is 1 unit. We can locate a score of 15 in quiz 2(halfway between 14 and 16). It also corresponds to a score of 15 in quiz 1. Thus, 1 student earned 15 marks in quiz 1 and 2
c) The equation of the line of best fit is written in the slope intercept form which is expressed as
y = mx + b
where
m = slope
b = y intercept
We would calculate the slope by applying the formula,
m = (y2 - y1)/(x2 - x1)
where
y1 and y2 are y coordinates of initial and final points on the line.
x1 and x2 are x coordinates of initial and final points on the line.
Picking points on the graph, we have
when x1 = 10, y1 = 8
when x2 = 16, y2 = 14
By substituting these values into the formula,
m = (14 - 8)/(16 - 10) = 6/6 = 1
We would find the y intercept by substituting m = 1, x = 10 and y = 8 into the slope intercept equation. We have
8 = 1 * 10 + b = 10 + b
b = 8 - 10
b = - 2
Substituting m = 1 and b = - 2 into the slope intercept equation, the equation of the line of best fit is
y = x - 2
The slope is 1 and since it is small, it tells us that for each score of 1 that a student gets in quiz 2, he would likely get a score of 1 in quiz 1.
Since the y intercept is negative, it doesn't make sense in the concept of the problem because a student cannot earn a negative score in any of the quizzes. The y intercept tells us that the student earned - 2 in quiz 2 and 0 in quiz 1