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Charra [1.4K]
3 years ago
13

PLEASE HELP!!

Mathematics
1 answer:
FrozenT [24]3 years ago
5 0

Answer:

Part A: 13 and 13, 26 and 26

Part B: yes

Step-by-step explanation:

if you distribute the 2, then it's the same expression as 6m+14. as a result, the values are all the same

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Step-by-step explanation:

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Step-by-step explanation:

well there is a very simple point that you have mistaken tho:

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5 0
2 years ago
There are 16 people in a running race. The first 3 people to finish the race will he given prizes. How many different ways can t
rodikova [14]

Answer : 3, 360 ways

Total number of people running the race = 16

We have three slot to be filled for first, second and third

Any person can take the first position irrespective of the arrangement

one out of the 16 people can come first

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7 0
1 year ago
based on the simulation, what is the probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to
Triss [41]

Using the binomial distribution, supposing that 0.3 of the callers have to wait more than 8 minutes to have their calls answered, it is found that there is a 0.3828 = 38.28% probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to have their calls answered.

For each caller, there are only two possible outcomes, either they have to wait more than 8 minutes to have their calls answered, or they do not. The probability of a caller having to wait more than 8 minutes is independent of any other caller, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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:

  • 10 callers, hence n = 10
  • Suppose that 0.3 of them have to wait more than 8 minutes, hence p = 0.3

The probability that <u>at most 2</u> of the next 10 callers will have to wait more than 8 minutes is:

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

Then

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

P(X = 0) = C_{10,0}.(0.3)^{0}.(0.7)^{10} = 0.0282

P(X = 1) = C_{10,1}.(0.3)^{1}.(0.7)^{9} = 0.1211

P(X = 2) = C_{10,2}.(0.3)^{2}.(0.7)^{8} = 0.2335

Then:

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0282 + 0.1211 + 0.2335 = 0.3828

0.3828 = 38.28% probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to have their calls answered.

A similar problem is given at brainly.com/question/25537909

3 0
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Answer:

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Step-by-step explanation:

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