Question

If a number x is chosen a random from the number -3, -2,-1, 0, 1, 2, 3. What is the probability that
${x }^{2} \leqslant 4$
​

Answer 1

Answer:

If a number x is chosen a random from the number -3, -2,-1, 0, 1, 2, 3. What is the probability ... - did not match any documents.

Answer 2

Answer:

$\sf \: \boxed{\sf \: Probability \: that \: {x}^{2} \leqslant 4 = \dfrac{5}{7} \: } \:$

Step-by-step explanation:

Given that, a number x is chosen a random from the number : -3, -2,-1, 0, 1, 2, 3.

Now, Number of outcomes = 7

Now, We have to find what is the probability that

${x }^{2} \leqslant 4$.

Now, Consider

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf {x}^{2} \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf - 3 & \sf 9\\ \\ \sf - 2 & \sf 4 \\ \\ \sf - 1 & \sf 1 \\ \\ \sf 0 & \sf 0\\ \\ \sf 1 & \sf 1\\ \\ \sf 2 & \sf 4\\ \\ \sf 3 & \sf 9\end{array}} \\ \end{gathered}$

So,

$\implies\sf \: Number\:of\:favourable\:outcomes = 5$

Now,

$\sf \: Probability \: that \: {x}^{2} \leqslant 4 \:$

$\sf \: = \: \dfrac{Number\:of\:favourable\:outcomes}{Number\:of\:possible\:outcomes}$

$\sf \: = \: \dfrac{5}{7}$

Hence,

$\implies\sf \: \boxed{\sf \: Probability \: that \: {x}^{2} \leqslant 4 = \dfrac{5}{7} \: }$