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Natali5045456 [20]
3 years ago
6

I need answer Immediately pls!!!!!!

Mathematics
1 answer:
Illusion [34]3 years ago
3 0

Given:

Total number of students = 27

Students who play basketball = 7

Student who play baseball = 18

Students who play neither sports = 7

To find:

The probability the student chosen at randomly from the class plays both basketball and base ball.

Solution:

Let the following events,

A : Student plays basketball

B : Student plays baseball

U : Union set or all students.

Then according to given information,

n(U)=27

n(A)=7

n(B)=18

n(A'\cap B')=7

We know that,

n(A\cup B)=n(U)-n(A'\cap B')

n(A\cup B)=27-7

n(A\cup B)=20

Now,

n(A\cup B)=n(A)+n(B)-n(A\cap B)

20=7+18-n(A\cap B)

n(A\cap B)=7+18-20

n(A\cap B)=25-20

n(A\cap B)=5

It means, the number of students who play both sports is 5.

The probability the student chosen at randomly from the class plays both basketball and base ball is

\text{Probability}=\dfrac{\text{Number of students who play both sports}}{\text{Total number of students}}

\text{Probability}=\dfrac{5}{27}

Therefore, the required probability is \dfrac{5}{27}.

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7 0
3 years ago
A cell phone company recently noticed a 26% sales decrease in the month of July compared to June. Let n represent June's
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Answer:

Option A. is correct.

Step-by-step explanation:

Given: A cell phone company recently noticed a 26\% sales decrease in the month of July compared to June such that n represent June's  sales.

Also, y represents July's sales.

To find: equation that can be used to determine the sales in July

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June's sales =n

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A single die is rolled. Find the probability of rolling an oddnumber or a number less than 6.
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Given

A single die is rolled.

To find the probability of rolling an odd number or a number less than 6.

Explanation:

It is given that,

A single die is rolled.

Then, the sample space is,

\begin{gathered} S=\lbrace1,2,3,4,5,6\rbrace \\ n(S)=6 \end{gathered}

Let A be the event of getting an odd number.

Then,

\begin{gathered} A=\lbrace1,3,5\rbrace \\ n(A)=3 \end{gathered}

Let B be the event of getting a number less than 6.

Then,

\begin{gathered} B=\lbrace1,2,3,4,5\rbrace \\ n(B)=5 \end{gathered}

Also,

\begin{gathered} A\cap B=\lbrace1,3,5\rbrace \\ n(A\cap B)=3 \end{gathered}

Therefore,

\begin{gathered} P(A\cup B)=P(A)+P(B)-P(A\cap B) \\ =\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A\cap B)}{n(S)} \\ =\frac{3}{6}+\frac{5}{6}-\frac{3}{6} \\ =\frac{3+5-3}{6} \\ =\frac{5}{6} \end{gathered}

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1 year ago
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