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Reply With Quote9/x > x resolves to:
9 > x^2 ===> -3 < x < 3
As long as 0<x<10
we have x={1,2} note that zero is neither positive nor negative.
probability = 2/9 = {1,2} / {1,2,3,4,5,6,7,8,9}
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A number, x is chosen at random from the set of positive integers less than 10. What is the probability that (9/x) > x?
(A) 1/5
(B) 2/9
(C) 1/3
(D) 2/3
(E) 7/9
SPOILER: Actual Answer : B
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9/x > x resolves to:
9 > x^2 ===> -3 < x < 3
As long as 0<x<10
we have x={1,2} note that zero is neither positive nor negative.
probability = 2/9 = {1,2} / {1,2,3,4,5,6,7,8,9}
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x^2 < 9 is true for x=1 and x=2 out of 9 possibilities...
-Amor
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Sorry, but could someone please try to explain again? I can't see how you come to 9 > x^2 and what does ^ mean?
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