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jek_recluse [69]
3 years ago
10

Decide if the situation involves​ permutations, combinations, or neither. Explain your reasoning. The number of ways a four dash

member committee can be chosen from 13 people. Does the situation involve​ permutations, combinations, or​ neither?
Choose the correct answer below.

a. Combination. The order of the committee members does not matter.

b. Permutation. The order of the committee members matters.

c. Neither A line of people is neither an ordered arrangement of objects, nor a selection of objects from a group of objects.
Mathematics
1 answer:
TEA [102]3 years ago
6 0

Answer: a. Combination. The order of the committee members does not matter.

Step-by-step explanation:

From the question we were only asked to look for a way to select four people from a 13 members committee the order for which it is selected wasn’t stated this makes it a perfect combination problem. Because permutations always have to do with order.

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There is no way to know.

Step-by-step explanation:

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At the end of a snow storm, Audrey saw there was a lot of snow on her front lawn.
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An equation that represents the depth of snow on Audrey's lawn, in inches, t hours after the snow stopped falling is S = 10 - 2t

In this question, there was a depth of 10 inches of snow on the lawn when the storm ended and then it started a melting at a rate of 2 inches per hour.

We need to write an equation for S in terms of t, snow representing the depth of snow on Audrey's lawn, in inches, t hours after the snow stopped falling.

So, the equation would be,

S = 10 - 2t

Therefore,  an equation that represents the depth of snow on Audrey's lawn, in inches, t hours after the snow stopped falling is S = 10 - 2t

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In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead. Construct 90% confiden
hram777 [196]

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(0.4958, 0.7422)

Step-by-step explanation:

Let p be the true proportion of water specimens that contain detectable levels of lead. The point estimate for p is \hat{p}=26/42=0.6190. The estimated standard deviation is given by \sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{(0.6190)(1-0.6190)/42}=0.0749. Because we have a large sample, the 90% confidence interval for p is given by 0.6190\pm z_{0.05}0.0749 where z_{0.05}=1.6448 is the value that satisfies that above this and under the standard normal density there is an area of 0.05. So, the confidence interval is 0.6190\pm (1.6448)(0.0749), i.e., (0.4958, 0.7422).

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