Difference between revisions of "2019 AMC 10A Problems/Problem 12"
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− | There are <math>365</math> values which means that the median is the <math>183</math>rd biggest value. Because there are <math>12</math> of each of the first <math>28</math> numbers, the median is <math>183/12=15.25</math>.The mean of these numbers is a little under <math>16</math> because there are only <math>7</math> <math>31</math>'s. The median of the modes is the median of <math>1</math> to <math>28</math>, yielding 14.5. Because <math>14.5<15.25<16</math>, the answer is < | + | There are <math>365</math> values which means that the median is the <math>183</math>rd biggest value. Because there are <math>12</math> of each of the first <math>28</math> numbers, the median is <math>183/12=15.25</math>.The mean of these numbers is a little under <math>16</math> because there are only <math>7</math> <math>31</math>'s. The median of the modes is the median of <math>1</math> to <math>28</math>, yielding 14.5. Because <math>14.5<15.25<16</math>, the answer is <math>B</math> |
-Lcz | -Lcz |
Revision as of 18:16, 9 February 2019
- The following problem is from both the 2019 AMC 10A #12 and 2019 AMC 12A #7, so both problems redirect to this page.
Problem
Melanie computes the mean , the median , and the modes of the values that are the dates in the months of . Thus her data consist of , , . . . , , , , and . Let be the median of the modes. Which of the following statements is true?
Solution
There are values which means that the median is the rd biggest value. Because there are of each of the first numbers, the median is .The mean of these numbers is a little under because there are only 's. The median of the modes is the median of to , yielding 14.5. Because , the answer is
-Lcz
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.