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Gennadij [26K]
3 years ago
11

If 40% of Jeanne's friends play kickball on weekends, what fraction of her friends don't play kickball?

Mathematics
2 answers:
Blababa [14]3 years ago
7 0

Answer:

3/5

Step-by-step explanation:

40% play kickball on weekends which would be 40/100=2/5, and the other 60% are the ones who don't play kickball so 60/100=6/10=3/5.

STatiana [176]3 years ago
3 0

the answer is (60% or 6/10 before simplifeing it 3/5 is the answer)

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13 quarters = 13 *0.25 = 3.25

3.60 - 3.25 = 0.35

0.35 = 0.25 +0.10

he has 1 dime and 14 quarters

Answer is B
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Julien was involved in an accident in which he merged into another car, damaging the front fender of the other car. Julien is fu
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Collision coverage pays for damage to your car resulting from a collision with another car, an object, such as a tree or telephone pole, or as a result of flipping over (note that collisions with deer are covered under comprehensive). It also covers damage caused by potholes.

Hope this helped. :)

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3 years ago
Find the solution to the system of equations.
Anestetic [448]

Answer:

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3 years ago
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It is estimated that 0.54 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal. Wha
stira [4]

Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.

<h3>What is the binomial distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 0.54% of the calls receive a busy signal, hence  p = 0.0054.
  • A sample of 1300 callers is taken, hence n = 1300.

The probability that at least 5 received a busy signal is given by:

P(X \geq 5) = 1 - P(X < 5)

In which:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

Then:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{1300,0}.(0.0054)^{0}.(0.9946)^{1300} = 0.0009

P(X = 1) = C_{1300,1}.(0.0054)^{1}.(0.9946)^{1299} = 0.0062

P(X = 2) = C_{1300,2}.(0.0054)^{2}.(0.9946)^{1298} = 0.0218

P(X = 3) = C_{1300,3}.(0.0054)^{3}.(0.9946)^{1297} = 0.0513

P(X = 4) = C_{1300,4}.(0.0054)^{4}.(0.9946)^{1296} = 0.0903

Then:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.

P(X \geq 5) = 1 - P(X < 5) = 1 - 0.1705 = 0.8295

0.8295 = 82.95% probability that at least 5 received a busy signal.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

6 0
2 years ago
A 2-column table with 6 rows. Column 1 is labeled Number of Items Made with entries 1,125, blank, blank, blank, blank, blank. Co
Nadya [2.5K]

Answer:

4,500 items are made every 4 hours.

5,625 items are made every 5 hours.

Step-by-step explanation:

when you multiply 1125 by for it will equal 4500 and when you multiply 1125 by 5 it will equal 5625

3 0
3 years ago
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