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expeople1 [14]
3 years ago
11

Please help me

Mathematics
1 answer:
Nataly_w [17]3 years ago
5 0

Jane is given a $100 gift to start and saves $35 a month from her allowance.

   After 1 month, Jane has saved

   After 2 months, Jane has saved

   After three months, Jane has saved

   and so on

In general, after x months Jane has saved

This means that it makes sense to represent the relationship between the amount saved and the number of months with one constant rate (in this case the constant rate is 35). It makes sense because the amount of money increases by $35 each month. Since the amount of increase is constant, we get constant rate. Also the initial amount is known ($100), so there is a possibility to write the equation of linear function representing this situation.

Step-by-step explanation:

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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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Ad libitum [116K]

390625

^if that’s what you meant in the picture cause your question was confusing

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