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VikaD [51]
4 years ago
8

A class consists of 60% men and 40% women. Blonde men compose 25% of the class, and blond women make up 20% of the class. If a s

tudent is chosen at random and is found to be a male, what is the probability that the student is blond?
Mathematics
2 answers:
Mice21 [21]4 years ago
4 0

Answer:

The probability that the student is blond and chosen be men is 65%.

Step-by-step explanation:

Given : A class consists of 60% men and 40% women. Blonde men compose 25% of the class, and blond women make up 20% of the class. If a student is chosen at random and is found to be a male.

To find : What is the probability that the student is blond?

Solution :  

Let the probability of men is 60% P(M)=0.6

The probability of women is 40% P(W)=0.4

Let B is the person blonde makeup.

The probability of blonde men is 25% P(B/M)=0.25

The probability of blonde women is 20% P(B/W)=0.20

We have to find the probability that the student is blonde and choosen is men i.e. P(M/B)

Applying Baye's Theorem,

P(M/B)=\frac{P(B/M)\times P(M)}{P(B/M)\times P(M)+P(B/W)\times P(W)}

P(M/B)=\frac{0.25\times 0.6}{0.25\times 0.6+0.20\times 0.4}

P(M/B)=\frac{0.15}{0.15+0.08}

P(M/B)=\frac{0.15}{0.23}

P(M/B)=0.65

Therefore, The probability that the student is blond and chosen be men is 65%.

anzhelika [568]4 years ago
3 0
<span>The probability that a selected male student is blond = 25/60 = 5/12 = 0.42</span>
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Answer:

a. 5 \frac{21}{40}

b. \frac{1}{42}

Step-by-step explanation:

a. \: 12 \frac{2}{5}  - 6 \frac{7}{8}

\frac{62}{5}  -  \frac{55}{8}

\frac{62}{8}  \times   \frac{8}{8}  -   \frac{55}{8}   \times  \frac{5}{5}

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\frac{221}{40}

5 \frac{21}{40}

b. \:  \frac{3}{14}  -  \frac{4}{21}

\frac{3}{14}  \times  \frac{3}{3}  -  \frac{4}{21}  \times  \frac{2}{2}

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\frac{1}{42}

Hope it is helpful....

5 0
3 years ago
Read 2 more answers
If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.
Pie
Given that f^{(n)}(0)=(n+1)!, we have for f(x) the Taylor series expansion about 0 as

f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n

Replace n+1 with n, so that the series is equivalent to

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Recall that for |x|, we have

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3 years ago
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Answer:

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3 0
3 years ago
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Answer:

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Step-by-step explanation:

Given that <CIO and <MAP are supplementary, therefore:

m<CIO + m<MAP = 180°

Let x be m<CIO

m<MAP = 5x

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Solve for x

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Divide both sides by 6

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The smaller angle = m<CIO = x = 30°

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