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mojhsa [17]
1 year ago
11

there are 6 studenjts in the class. one of them is to be selcted as the best studen and another student is to be selceted as a r

unner-up. how mant different ways can this be done?
Mathematics
1 answer:
Zarrin [17]1 year ago
5 0

As per the combination method, there are 15 different ways are possible for this situation.

Combination method:

Combination method also known as NCR is a formula of combination and arrangement, where the object's order does not matter so much.

Given,

Here we have to find the number of ways to be selected as the best student and another student is to be selected as a runner-up among the 6 students.

Here we know that that the value of

Total number of students = 6

Best student = 1

Runner - up = 1

So, as per the NCR formula, the value of n = 6 and the value of r = 2

Then the possible combinations are calculated as,

=> ⁶C₂ = 6! / 2!(6 - 2)!

=> ⁶C₂ = 6!/2!4!

=>⁶C₂ = 15

Therefore, the resulting number of ways is 15.

To know more about Combination method here.

brainly.com/question/28998705

#SPJ4

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3 years ago
At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro
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Answer:

A sample of 1068 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

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So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of \pi, so we use the worst case scenario, which is \pi = 0.5

Then

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.03\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.03}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}

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Rounding up

A sample of 1068 is needed.

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