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3 years ago

Pls answer fast i need help

Mathematics

2 answers:

juin [17]3 years ago

6 0

Do it by your self ok ???

Mamont248 [21]3 years ago

4 0

Answer:

Step-by-step explanation:

x^(a-(-b))(a-b).x^(b-(-c))(b-c).x^(c-(+a))(c-a)

x^(a+b)(a-b).x^(b+c)(b-c).x^(c+a)(c-a)

x^(a2-b2).x^(b2-c2).x^(c2-a2)

x^a2-b2+b2-c3+c2-a2

x^0

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at a particular high school, 42% of the students participate in sports and 25% of the students participate in drama. if 53% of t

baherus [9]

The probability that a student participates in both sports and drama is $0.14$ .

<h3>What is the formula for P(AUB), where A and B are any two events?</h3>

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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