Difference between revisions of "1992 AJHSME Problems/Problem 11"
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The total frequency is <math> 50+60+40+60+40=250 </math>, with the blue frequency of <math> 60 </math>. Therefore, the precentage that preferred blue is <math> \frac{60}{250}=\boxed{\text{(B)}\ 24\%} </math>. | The total frequency is <math> 50+60+40+60+40=250 </math>, with the blue frequency of <math> 60 </math>. Therefore, the precentage that preferred blue is <math> \frac{60}{250}=\boxed{\text{(B)}\ 24\%} </math>. | ||
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+ | ==See Also== | ||
+ | {{AJHSME box|year=1992|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Latest revision as of 23:09, 4 July 2013
Problem
The bar graph shows the results of a survey on color preferences. What percent preferred blue?
Solution
The total frequency is , with the blue frequency of . Therefore, the precentage that preferred blue is .
See Also
1992 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 |
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