Difference between revisions of "2010 AMC 8 Problems/Problem 4"
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Putting the numbers in numerical order we get the list <math>0,0,1,2,3,3,3,4.</math> | Putting the numbers in numerical order we get the list <math>0,0,1,2,3,3,3,4.</math> | ||
The mode is <math>3.</math> The median is <math>\frac{2+3}{2}=2.5.</math> The average is <math>\frac{0+0+1+2+3+3+3+4}{8}=\frac{16}{8}=2.</math> The sum of all three is <math>3+2.5+2=\boxed{\textbf{(C)}\ 7.5}</math> | The mode is <math>3.</math> The median is <math>\frac{2+3}{2}=2.5.</math> The average is <math>\frac{0+0+1+2+3+3+3+4}{8}=\frac{16}{8}=2.</math> The sum of all three is <math>3+2.5+2=\boxed{\textbf{(C)}\ 7.5}</math> | ||
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+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/51K3uCzntWs?t=209 | ||
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+ | ~ pi_is_3.14 | ||
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+ | ==Video by MathTalks== | ||
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+ | https://youtu.be/EEbksvfujhk | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=3|num-a=5}} | {{AMC8 box|year=2010|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:00, 5 February 2023
Problem
What is the sum of the mean, median, and mode of the numbers ?
Solution
Putting the numbers in numerical order we get the list The mode is The median is The average is The sum of all three is
Video Solution by OmegaLearn
https://youtu.be/51K3uCzntWs?t=209
~ pi_is_3.14
Video by MathTalks
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.