PLEASE HELP I’LL GIVE BRAINLIEST!

1 answer:

3 0

Here the first one, One goal in solving an equation is to have only variables on one side of the equal sign and numbers on the other side of the equal sign. The other goal is to have the number in front of the variable equal to one. The strategy for getting the variable by itself with a coefficient of 1 involves using opposite

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PLEASE HELP I WILL MARK YOU THE BRAINIEST On the same grid paper above, transform your figure ABC using the rule (2x,2y)

xeze [42]

Answer:

move each dot up two lines and right two lines it should transfer the image to the right spot

Step-by-step explanation:

6 0

2 years ago

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:

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