If $A$ and $B$ are any two events, then the probability of the joint event $(A\cup B)$ is given by the following formula: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Given that 42% of the students participate in sports and 25% of the students participate in drama and 53% of the students participate in either sports or drama.

Suppose $S$ denotes that "a student participates in sports" and $D$ denotes that "a student participates in drama".
So, we have $P(S)=\frac{42}{100}=0.42$ , $P(D)=\frac{25}{100}=0.25$ , $P(S\cup D)=\frac{53}{100}=0.53$ .

We want to find the probability that a student participates in both sports and drama i.e., we want to find $P(S\cap D)$ .

By the above formula, we obtain:

$P(S\cup D)=P(S)+P(D)-P(S\cap D)\\\Longrightarrow P(S\cap D)=P(S)+P(D)-P(S\cup D)\\\Longrightarrow P(S\cap D)=0.42+0.25-0.53\\\therefore P(S\cap D)=0.14=\frac{14}{100}$

Therefore, the probability that a student participates in both sports and drama is $0.14$ .

To learn more about probability, refer: brainly.com/question/24756209

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6 0

2 years ago

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Solve the equation without the use of a calculator. please Help!!! Thank you!!!<br> 12x^3-30x=5-2x^2

nata0808 [166]

So, since we have a cubic equation with 4 terms, the first thing we should try is factoring by grouping, so:

$12x^3+2x^2-30x+5 =0 \implies \\ 2x^2(6x+1)-5(6x+1)=0 \implies \\ (2x^2-5)(6x+1) = 0$

Now that we've factored our equation, we can use ZPP and break it up:

$2x^2-5 = 0 \implies \\ 2x^2=5 \implies\\ x^2=\frac{5}{2} \implies\\ x=\pm\frac{\sqrt{5}}{\sqrt{2}}=\pm\frac{\sqrt{10}}{2} \\ \\ 6x+1 =0 \implies \\ 6x=-1 \implies\\ x=\frac{-1}{6}$

So, our solutions are:

$x\in \{\frac{\sqrt{10}}{2},-\frac{\sqrt{10}}{2},-\frac{1}{6}\}$

4 0

3 years ago