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(A) We are given the inequality $\frac{(x-10)(x-50)}{(x-30)} \ge 0$. Using the wavy curve method (sign scheme) for the expression $f(x) = \frac{(x-10)

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$0.71$

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(A) We are given the inequality $\frac{(x-10)(x-50)}{(x-30)} \ge 0$. Using the wavy curve method (sign scheme) for the expression $f(x) = \frac{(x-10)(x-50)}{(x-30)}$,the critical points are $x = 10, 30, 50$. The sign of $f(x)$ in different intervals is: - For $x - For $10 \le x - For $30 - For $x \ge 50$: $f(x) \ge 0$ Since $x \in \{1, 2, \dots, 100\}$,we look for integers $x$ in the intervals $[10, 30)$ and $[50, 100]$. In the interval $[10, 30)$,the integers are $\{10, 11, \dots, 29\}$. The number of such integers is $29 - 10 + 1 = 20$. In the interval $[50, 100]$,the integers are $\{50, 51, \dots, 100\}$. The number of such integers is $100 - 50 + 1 = 51$. The total number of favorable outcomes is $20 + 51 = 71$. The total number of possible outcomes is $100$. Therefore,$P(A) = \frac{71}{100} = 0.71$.

$A$ number $x$ is chosen at random from the set $\{1, 2, 3, 4, \dots, 100\}$. Define the event: $A =$ the chosen number $x$ satisfies $\frac{(x - 10)(x - 50)}{(x - 30)} \ge 0$. Then $P(A)$ is

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