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(D) The given inequation is $x + \frac{10}{x} \leq 11$. Since $x \in \{1, 2, 3, \ldots, 50\}$, $x$ is always positive. Multiplying by $x$, we get $x^2

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$\frac{1}{5}$

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(D) The given inequation is $x + \frac{10}{x} \leq 11$. Since $x \in \{1, 2, 3, \ldots, 50\}$, $x$ is always positive. Multiplying by $x$, we get $x^2 + 10 \leq 11x$, which simplifies to $x^2 - 11x + 10 \leq 0$. Factoring the quadratic, we get $(x - 1)(x - 10) \leq 0$. This inequality holds for $1 \leq x \leq 10$. The integers satisfying this condition are $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. There are $10$ such integers. The total number of possible outcomes is $50$. The probability is $\frac{10}{50} = \frac{1}{5}$.

If a number $x$ is drawn randomly from the set of numbers $\{1, 2, 3, \ldots, 50\}$, then the probability that the number $x$ satisfies the inequation $x + \frac{10}{x} \leq 11$ is

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