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10) can fulfill (x

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10) can fulfill (x

by sanju09 » Thu Apr 09, 2009 4:24 am
What is the probability that a number randomly selected from (-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10) can fulfill (x- 5) (x + 10) (2 x - 5) = 0?
A. 1/12
B. 1/6
C. 1/4
D. 1/3
E. 1/2




OA B
Last edited by sanju09 on Fri Apr 10, 2009 4:41 am, edited 1 time in total.
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by moutar » Thu Apr 09, 2009 6:21 am
(x- 5) (x + 10) (2 x - 5) = 0

Solutions:
x - 5 = 0 -> x = 5
x + 10 = 0 -> x = -10
2x - 5 = 0 -> x = 2.5

2 of the solutions are in the set of 12.

Therefore probability = 2/12 = 1/6 B.
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