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probablity for an eq [#permalink]
25 Oct 2010, 06:43

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Difficulty:

35% (medium)

Question Stats:

73% (01:53) correct
27% (00:18) wrong based on 11 sessions

{-10,-6,-5,-4,-2.5,-1,0,2.5,4,6,7,10}

A number is selectedat random from the set above. what is the probablitiy that the number selected will be a solution of the equation (x-5)(x+10)(2x-5)=0? 1/12 1/6 1/4 1/3 1/2

Re: probablity for an eq [#permalink]
25 Oct 2010, 06:50

Expert's post

vanidhar wrote:

{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}

A number is selectedat random from the set above. what is the probablitiy that the number selected will be a solution of the equation (x-5)(x+10)(2x-5)=0? 1/12 1/6 1/4 1/3 1/2

Solutions of the equation \((x-5)(x+10)(2x-5)=0\) are: \(x=5\), \(x=-10\) and \(x=2.5\).

There are total of 12 numbers in the set {-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10} and it contains 2 roots: -10 and 2,5, so the probablitiy that the number selected will be a solution of the given equation is \(P=\frac{2}{12}=\frac{1}{6}\).

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