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Updated on: 28 Nov 2025, 04:20
Originally posted by Sajjad1994 on 20 Jan 2020, 08:30.
Last edited by Bunuel on 28 Nov 2025, 04:20, edited 3 times in total.
Last edited by Bunuel on 28 Nov 2025, 04:20, edited 3 times in total.
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Dropdown 1: -1.3
Dropdown 2: 100% greater
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
49% (02:57) correct
51%
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wrong
based on 469
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At a certain archery school, each of five students shot a single arrow at the end of each day of training, as well as one arrow before the first day of training. The graph above is a scatterplot, in which each of the 30 points represents the distance from the center of the target to each student’s arrow and the number of days the student had been in training at the time the arrow was shot. The solid line is the regression line.
Use the drop-down menus to fill in the blanks in each of the following statements based on the information given by the graph.
The slope of the regression line is closest to .
The number of students within 11 inches of the center of the target is after day 2 of training than before any training.
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Official Answer
Dropdown 1: -1.3
Dropdown 2: 100% greater
20 Jan 2020, 08:49
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1) -1.3,since the two points on the regression line are (1,10) & (4,6) so the slope will be : (6-10)/(4-1)= -4/3= -1.33
2)After day 2, number of candidates who have targets within 11 inches = 4,while before any training the number is 2, so it's a 100% increment
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2)After day 2, number of candidates who have targets within 11 inches = 4,while before any training the number is 2, so it's a 100% increment
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30 Apr 2024, 11:50
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Official Explanation
The graph is clearly labeled, so there’s no reason to read the text blurb. The first question asks for the slope of the regression line. Open the drop-down box and review the answer choices. The line’s slope is negative, so eliminate (D) and (E). The line clearly passes through points (1, 10) and (4, 6), so the slope of the line is approximately equal to \(\frac{y-y_{1}}{x-x_{1}}= \frac{6-10}{4-1}\), or \(\frac{-4}{3}\) . Expressed as a decimal, \(\frac{-4}{3}\) is approximately equal to -1.3.
The correct answer is (B).
The question asks for the percent by which the number of students within 11 inches of the target’s center after day 2 is greater or less than it was before any training. Open the drop-down box and review the answer choices. More students were within 11 inches of the target after 2 days of training, so eliminate (A) and (B). Before any training, 2 students were within 11 inches of the target’s center. After 2 days of training, 4 students were within 11 inches of the target’s center. Use the percent change formula, (difference/original) × 100, to calculate the percent increase, which is equal to \(\frac{4-2}{2} × 100\), or 100%.
The correct answer is (D).
The graph is clearly labeled, so there’s no reason to read the text blurb. The first question asks for the slope of the regression line. Open the drop-down box and review the answer choices. The line’s slope is negative, so eliminate (D) and (E). The line clearly passes through points (1, 10) and (4, 6), so the slope of the line is approximately equal to \(\frac{y-y_{1}}{x-x_{1}}= \frac{6-10}{4-1}\), or \(\frac{-4}{3}\) . Expressed as a decimal, \(\frac{-4}{3}\) is approximately equal to -1.3.
The correct answer is (B).
The question asks for the percent by which the number of students within 11 inches of the target’s center after day 2 is greater or less than it was before any training. Open the drop-down box and review the answer choices. More students were within 11 inches of the target after 2 days of training, so eliminate (A) and (B). Before any training, 2 students were within 11 inches of the target’s center. After 2 days of training, 4 students were within 11 inches of the target’s center. Use the percent change formula, (difference/original) × 100, to calculate the percent increase, which is equal to \(\frac{4-2}{2} × 100\), or 100%.
The correct answer is (D).
