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Yanka [14]
2 years ago
7

According to a survey of adults, 64 percent have money in a bank savings account. If we were to survey 50 randomly selected adul

ts, find the mean number of adults who would have bank savings accounts.
Mathematics
1 answer:
zheka24 [161]2 years ago
6 0

Answer:

The mean number of adults who would have bank savings accounts is 32.

Step-by-step explanation:

For each adult surveyed, there are only two possible outcomes. Either they have bank savings accounts, or they do not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

E(X) = np

In this problem, we have that:

p = 0.64

If we were to survey 50 randomly selected adults, find the mean number of adults who would have bank savings accounts.

This is E(X) when n = 50.

So

E(X) = np = 50*0.64 = 32

The mean number of adults who would have bank savings accounts is 32.

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All we have to do here is divide 4 by 2

4 ÷ 2 = 2

Therefore, the artist can paint 2 paintings an hour making the unit rate, 2 paintings per hour.

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irina [24]

Answer:

the expected value is 300

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Jeremy wants to give a 20% gratuity to his cab driver. His fare was $35.50. What is the total amount he will pay the driver?
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The function A(b) relates the area of a trapezoid with a given height of 10 and
Natali5045456 [20]

The function of the trapezoid area is:

A(x)=(B+b)*h/2

Where B and b are the bases and h is the height.

With the given data: h=10 B and b =7 and x (it may vary which one is bigger)

-----------

So that function becomes:

A(x)=(7+x)*10/2

A(x)=(7+x)*5

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So if you want the inverse function, you have to operate to find x:

A(x)/5=7+x

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6 0
2 years ago
Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The
abruzzese [7]

The question is incorrect.

The correct question is:

Three TAs are grading a final exam.

There are a total of 60 exams to grade.

(c) Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The TAs grade at different rates, so the first TA will grade 25 exams, the second TA will grade 20 exams and the third TA will grade 15 exams. How many ways are there to distribute the exams?

Answer: 60!/(25!20!15!)

Step-by-step explanation:

The number of ways of arranging n unlike objects in a line is n! that is ‘n factorial’

n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1

The number of ways of arranging n objects where p of one type are alike, q of a second type are alike, r of a third type are alike is given as:

n!/p! q! r!

Therefore,

The answer is 60!/25!20!15!

6 0
1 year ago
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