Difference between revisions of "2010 AMC 8 Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | Putting the numbers in numerical order we get the list 0,0,1,2,3,3,3,4 | + | Putting the numbers in numerical order we get the list <math>0,0,1,2,3,3,3,4.</math> |
| − | The mode is 3 The median is | + | 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> |
==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 14:22, 15 August 2021
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 
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. 
