An integer x is chosen from 1 to 50. Then the | vedclass

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(A) The given inequality is $x + \frac{336}{x} \le 50$.Since $x$ is a positive integer from $1$ to $50$,we can multiply by $x$ without changing the in

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$7/10$

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(A) The given inequality is $x + \frac{336}{x} \le 50$.Since $x$ is a positive integer from $1$ to $50$,we can multiply by $x$ without changing the inequality sign:$x^2 + 336 \le 50x$$x^2 - 50x + 336 \le 0$Factoring the quadratic expression,we get:$(x - 8)(x - 42) \le 0$This inequality holds for $x \in [8, 42]$.Since $x$ is 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 probability is $P = \frac{35}{50} = \frac{7}{10}$.

An integer $x$ is chosen from $1$ to $50.$ Then the probability that $x + \frac{336}{x} \le 50$ is

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