Official Explanation
Statement 1: –$3,500. The problem first specifies that it is talking about only the branches that had an increase in variable costs during the second half of the year. This description corresponds to those points with a positive y-coordinate (i.e. above the x-axis).
There are 9 branches above the x-axis. From this group only, find the one with the median change in cost during the first half of the year (the x-axis). The median is the 5th data point from either end of the x-axis (the physical middle). Count over 5 from the left or 5 from the right; the median branch is the point at approximately (-3.5, 19). Because the change in variable cost during the earlier period is given by the x-coordinate (in thousands), the median branch experienced a change of \(-3.5 × $1000 = -$3,500.\)
Statement 2: 47%. To find branches with a net decrease in costs, it will help to understand what each quadrant represents.
Quadrant I (upper right): These branches had an increase over both periods (3 branches fall into this category).
Quadrant II (upper left): These branches had a decrease during the first period but an increase during the second (6 branches).
Quadrant III (lower left): These branches had a decrease during both periods (1 branch).
Quadrant IV (lower right): These branches had an increase during the first period but a decrease during the second (5 branches).
None of the branches that have an increase during both periods will have a net overall decrease, so ignore the branches in Quadrant I (3 branches). All branches that have a decrease during both periods will have a net overall decrease, so count all branches in Quadrant III (1 branch).
Now, address the more complicated situations—an increase in one period but a decrease in the other. To arrive at a net overall decrease, you need a larger dollar decrease than increase. For Quadrant II (decrease in first period but increase in second), the x-coordinate (period 1) must have a larger magnitude than the y-coordinate (period 2). This is the case for 3 points: (-7, 5.5), (-15.5, 5.5), and (-16, 0.5). For each of these points, the decrease outweighs the increase (do not calculate the actual values).
Finally, for Quadrant IV (decrease in the second period but increase in the first), the y-coordinate (period 2) must have a larger magnitude than the x-coordinate (period 1). This is the case for 3 points (2.5, -4.5), (0.5, -7.5), and (11, -17). For each of these points, the decrease again outweighs the increase.
A total of 1 + 3 + 3 = 7 branches out of the 15 have a net decrease. 7/15 is slightly less than 1/2 or 50%; 47% is the best estimate.
Alternatively, if you feel very comfortable with coordinate plane problems, you could find all the points satisfying the expression x + y < 0 (i.e. the sum of the changes is negative). Rearrange in order to graph: y < –x. Where does that line appear on the coordinate plane? This is the equivalent of graphing the line y = –x and then shading underneath that line. Any points underneath the line satisfy that inequality.
There are 7 points located below that line out of 15 total branches, so the best estimate is a little bit less than 1/2, or approximately 47%.
Answer: B and B
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