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Vedmedyk [2.9K]
3 years ago
7

Need answer 25 and answer 27

Mathematics
1 answer:
UNO [17]3 years ago
8 0

\\ \bull\tt\dashrightarrow \dfrac{-3}{8}x-20+2x>6

\\ \bull\tt\dashrightarrow \dfrac{-3}{8}x+2x=6+20=26

\\ \bull\tt\dashrightarrow \dfrac{-3+16}{8}x=26

\\ \bull\tt\dashrightarrow \dfrac{13}{8}x=26

\\ \bull\tt\dashrightarrow x=26\times \dfrac{8}{13}

\\ \bull\tt\dashrightarrow x=8(2)=16

#27

\\ \bull\tt\dashrightarrow 0.5x-4-2x\leqslant 2

\\ \bull\tt\dashrightarrow 0.5x-2x\leqslant=2+4=6

\\ \bull\tt\dashrightarrow -1.5x\leqslant 6

\\ \bull\tt\dashrightarrow x\leqslant\dfrac{6}{-1.5}

\\ \bull\tt\dashrightarrow x\leqslant 4

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At a local company, 15% of the employees are women. every day, 9% of them bring their lunch to work, while only 3% of the men br
fenix001 [56]

Answer:

a) 0.3462 = 34.62% that a randomly selected employee is a woman given that the person brings their lunch to work.

b) 0.09 = 9% probability that a randomly selected employee brings their lunch to work given that person is a woman.

c) 0.3462 = 34.62% that a randomly selected employee is a woman given that the person brings their lunch to work.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

Questions a/c:

Questions a and c are the same, so:

Event A: Brings lunch to work.

Event B: Is a woman.

Probability of a person bringing lunch to work:

9% of 15%(woman)

3% of 100 - 15 = 85%(man). So

P(A) = 0.09*0.15 + 0.03*0.85 = 0.039

Probability of a person bringing lunch to work and being a woman:

9% of 15%, so:

P(A \cap B) = 0.09*0.15

Desired probability:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.09*0.15}{0.039} = 0.3462

0.3462 = 34.62% that a randomly selected employee is a woman given that the person brings their lunch to work.

Question b:

Event A: Woman

Event B: Brings lunch

15% of the employees are women.

This means that P(A) = 0.15

Probability of a person bringing lunch to work and being a woman:

9% of 15%, so:

P(A \cap B) = 0.09*0.15

Desired probability:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.09*0.15}{0.15} = 0.09

0.09 = 9% probability that a randomly selected employee brings their lunch to work given that person is a woman.

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