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(B) The set of integers is $S = \{x \in Z \mid -25 \leq x \leq 25, x 
eq 0\}$. The total number of elements is $50$.We need to solve the inequality $

Vedclass · Accepted Answer

$\frac{2}{5}$

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(B) The set of integers is $S = \{x \in Z \mid -25 \leq x \leq 25, x 
eq 0\}$. The total number of elements is $50$.We need to solve the inequality $x + 6 \leq \frac{135}{x}$.Rearranging the inequality,we get $\frac{x^2 + 6x - 135}{x} \leq 0$.Factoring the numerator,we have $\frac{(x + 15)(x - 9)}{x} \leq 0$.Using the sign scheme method,the solution set for the inequality is $x \in (-\infty, -15] \cup (0, 9]$.Intersecting this with the given set $S$,we get $x \in [-25, -15] \cup \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.The number of integers in $[-25, -15]$ is $11$,and the number of integers in ${1, 2, 3, 4, 5, 6, 7, 8, 9}$ is $9$.Total favorable outcomes $= 11 + 9 = 20$.Probability $= \frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{20}{50} = \frac{2}{5}$.

$A$ non-zero integer $x$ is selected randomly from the set of integers $\{x \in Z \mid -25 \leq x \leq 25, x \neq 0\}$. The probability that $x + 6 \leq \frac{135}{x}$ is

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