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What is the possibility such that the inequality x^2 + b > ax is true, when a=32.4 and b=76.5 and x∈[0,30].(a) 1.91(b) 4.3(c) 2.94(d) 6.1The question was asked during an interview for a job.I need to ask this question from Geometric Probability topic in chapter Discrete Probability of Discrete Mathematics

Answer» CORRECT choice is (a) 1.91

To ELABORATE: x2+76.5>32.4x is EQUIVALENT to x^2−32.4x+76.5 > 0. By completing the square, x^2 − 32.4x + 266.44 – 266.44 + 76.5 = (x−16.2)^2 − 189.94>0, which is the same as (x−16.2)^2 > 189.94,which implies that either x−16.2 > 13.78 ⇒ x > 29.98, or x−16.2 < −2.42 ⇒ x< 13.78. Assume that it is a uniform DISTRIBUTION. So, the PROBABILITY that x > 29.98 is 30 − 29.98 = 0.02 and the probability that x < 189.94 is 1.89. The desired probability is 0.02 + 1.89 = 1.91.


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