In a class of 30 students, 6 have a brother and 12 have a sister. There are 14 students who do not have any siblings. What is the probability that a student chosen randomly from the class has a brother and a sister?

Answers

Answer 1

Answer:

32 siblings from the class theres a possibility that some students had more siblings and not everyone has one

Step-by-step explanation:

Answer 2

The probability that a student chosen randomly from the class has a brother and a sister is [tex]\frac{1}{15}[/tex].

What is probability?

The chance that something will happen, or how likely it is that an event will occur.

Formula to calculate probability

probability of an event = [tex]\frac{number of favorable outcomes}{Total number of outcomes}[/tex]

According to the given question

we have total 30 students.

6 have a brother

12 have a sister

and 14 who don't have any siblings

If we se the provided Venn diagram

the number of students who have only sister is 10

and the number of student who have only brother is 4.

because, 6+12+14 = 32

but we have only 30 students

⇒ 2 students has a brother and a sister.

therefore,

the probability that a student chosen randomly from the class has a brother and a sister = [tex]\frac{number of students as a brother and a sister}{total number of students}[/tex]= [tex]\frac{2}{30}[/tex] =[tex]\frac{1}{15}[/tex]

Hence, the probability that the chosen student from the class has a brother and sister is [tex]\frac{1}{15}[/tex].

Learn more about probability here:

https://brainly.com/question/11234923

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In A Class Of 30 Students, 6 Have A Brother And 12 Have A Sister. There Are 14 Students Who Do Not Have

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Answers

Answer:

[tex]f_{ave}[/tex] = [tex]{\frac{10 }{3\pi }[/tex]

Step-by-step explanation:

To find - Find the average value fave of the function f on the given interval. f(x) = 5 sec²(x/6), [0, 3[tex]\pi[/tex]/2]

Proof -

We know that,

Average value of f in the interval [a, b] is -

[tex]f_{ave}[/tex] = [tex]\frac{1}{b - a}\int\limits^b_a {f(x)} \, dx[/tex]

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Here a = 0, b = [tex]\frac{3\pi }{2}[/tex], f(x) = [tex]5sec^{2}(\frac{x}{6} )[/tex]

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[tex]f_{ave}[/tex] = [tex]\frac{1}{\frac{3\pi }{2} - 0}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx[/tex]

= [tex]{\frac{2 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx[/tex]

= [tex]{\frac{10 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {sec^2 ({\frac{x}{6} }) } \, dx[/tex]

= [tex]{\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) }|^{\frac{3\pi }{2} }_0 \, ][/tex]

= [tex]{\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) - tan({0) } \, ][/tex]

= [tex]{\frac{10 }{3\pi }}[ {1 - 0 } ][/tex]

= [tex]{\frac{10 }{3\pi }[/tex]

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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