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 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
==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}} | ||
Revision as of 22:02, 25 December 2022
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
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. 
