An integer x is chosen at random from 1 to 50 | vedclass

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(A) Given the inequality $x + \frac{336}{x} \leq 50$. Since $x$ is a positive integer,we can multiply by $x$ without changing the inequality sign: $x^

Vedclass · Accepted Answer

$\frac{7}{10}$

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(A) Given the inequality $x + \frac{336}{x} \leq 50$. Since $x$ is a positive integer,we can multiply by $x$ without changing the inequality sign: $x^2 + 336 \leq 50x$. Rearranging the terms,we get $x^2 - 50x + 336 \leq 0$. Factoring the quadratic expression,we find $(x - 8)(x - 42) \leq 0$. This inequality holds for $x$ in the interval $[8, 42]$. Since $x$ must be an integer,the possible values for $x$ are ${8, 9, 10, \dots, 42}$. The number of such integers is $42 - 8 + 1 = 35$. The total number of integers from $1$ to $50$ is $50$. Therefore,the required probability is $\frac{35}{50} = \frac{7}{10}$.

An integer $x$ is chosen at random from $1$ to $50$. The probability that $x + \frac{336}{x} \leq 50$ is

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