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If the set S consists of all nonnegative integers less than 7, what [#permalink]
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08 Jul 2015, 03:32
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Re: If the set S consists of all nonnegative integers less than 7, what [#permalink]
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08 Jul 2015, 04:08
Bunuel wrote: If the set S consists of all nonnegative integers less than 7, what is the probability that a randomly selected integer x of set S will satisfy the inequality 3 > x^2/3?
A. 1/7 B. 2/7 C. 3/7 D. 4/7 E. 1
Kudos for a correct solution. Set, S = {0, 1, 2, 3, 4, 5, 6} [All NonNegative Integers]3 > x^2/3 i.e. x^2 < 3*3 i.e. x^2 < 9Which is true for the values of x = 0, 1 and 2 out of 7 values available in Set S i.e. Probability = Favourable Outcome / Total Outcome = 3/7 Answer: Option C
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Re: If the set S consists of all nonnegative integers less than 7, what [#permalink]
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08 Jul 2015, 04:44
Bunuel wrote: If the set S consists of all nonnegative integers less than 7, what is the probability that a randomly selected integer x of set S will satisfy the inequality 3 > x^2/3?
A. 1/7 B. 2/7 C. 3/7 D. 4/7 E. 1
Kudos for a correct solution. Non negative numbers less than 7 will be = {0,1,2,3,4,5,6} Given inequality : x^2<9 > (x+3)(x3)<0 > 3<x<3. Thus x can assume 0,1,2 as the values. Thus the probability = 3/7, C is the correct answer.



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Re: If the set S consists of all nonnegative integers less than 7, what [#permalink]
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08 Jul 2015, 06:56
Numbers satisfying the criteria are 0, 1, and 2, so 3/7. (C)
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Re: If the set S consists of all nonnegative integers less than 7, what [#permalink]
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08 Jul 2015, 22:01
Bunuel wrote: If the set S consists of all nonnegative integers less than 7, what is the probability that a randomly selected integer x of set S will satisfy the inequality 3 > x^2/3?
A. 1/7 B. 2/7 C. 3/7 D. 4/7 E. 1
Kudos for a correct solution. Set S is set of NONNEGATIVE INTEGER less than 7 , S = {0,1,2,3,4,5,6,7} Total Number of element in set S = 7 We need X such that 3 > x^2/3 or 9 > x^2 Only three number from set S satisfy this condition , those are 0,1,2 Hence Required Probability = Number of Element satisfying condition / Total number of element = 3/7 Hence Option C should be correct.



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Re: If the set S consists of all nonnegative integers less than 7, what [#permalink]
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13 Jul 2015, 02:27
Bunuel wrote: If the set S consists of all nonnegative integers less than 7, what is the probability that a randomly selected integer x of set S will satisfy the inequality 3 > x^2/3?
A. 1/7 B. 2/7 C. 3/7 D. 4/7 E. 1
Kudos for a correct solution. 800score Official Solution:We are told that S = {0, 1, 2, 3, 4, 5, 6}, that is, all integers from 0 to 6. There are 7 integers in the set. Remember that zero is an integer that is neither negative nor positive. Now, the expression can be simplified by multiplying both sides by 3. We can do this since the 3 in the denominator is a positive number. Remember: When you multiply or divide both sides of an inequality by a negative number, you must change the direction of the inequality. The expression is simplified to 9 > x^2. The only numbers in S that can be squared and satisfy this condition are 0, 1, and 2. Therefore, 3 of the 7 integers in the set will satisfy the condition, and the desired probability is 3/7. The correct answer is choice (C).
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Re: If the set S consists of all nonnegative integers less than 7, what [#permalink]
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13 Oct 2017, 09:28
Bunuel wrote: If the set S consists of all nonnegative integers less than 7, what is the probability that a randomly selected integer x of set S will satisfy the inequality 3 > x^2/3?
A. 1/7 B. 2/7 C. 3/7 D. 4/7 E. 1 We see that set S = {0, 1, 2, 3, 4, 5, 6}. Let’s test each number: 3 > 0^2/3 = 0? yes 3 > 1^2/3 = 1/3? yes 3 > 2^2/3 = 4/3? yes 3 > 3^2/3 = 9/3 = 3? no Since 3 of 7 numbers (namely, 0, 1, and 2) satisfy the inequality, the probability of getting these 3 numbers is 3/7. Answer: C
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