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Ira Lisetskai [31]
3 years ago
14

IF RIGHT I WILL MARK BRANILIEST! HELP!

Mathematics
1 answer:
notka56 [123]3 years ago
6 0

Answer:

D

Step-by-step explanation:

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And the answer to the circumference of a circle with a radius of 177 is approximately:

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You have to answer essay questions for an exam. There are essays to choose from. How many different groups of essays could you p
Lynna [10]

Answer:

15.

Step-by-step explanation:

Please consider the complete question.

You have to answer 2 essay questions for an exam. There are 6 essays to choose from. How many different groups of 2 essays could you possibly choose?

We will use combinations formula to solve our given problem.

The number of combination of n terms r terms chosen at a time is given by formula:

C(n,r)=\frac{n!}{r!(n-r)!}

C(6,2)=\frac{6!}{2!(6-2)!}

C(6,2)=\frac{6*5*4!}{2*1*4!}

C(6,2)=\frac{6*5}{2}

C(6,2)=3*5

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Therefore, there are 15 different groups of essays could you possibly choose.

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3 years ago
19 diameter of a circle in<br> Area<br> Circumference
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Answer:

59.66

Step-by-step explanation:

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What is 2.85 to the nearest tenth?
Maru [420]

The answer is 2.9. You would round up since there is a 5

6 0
4 years ago
Read 2 more answers
Question 15
larisa86 [58]

Answer:

The correct option is (C).

Step-by-step explanation:

According to the Central Limit Theorem if we have n unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample means is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 3.5

<em>σ</em> = 1.7078

<em>n</em> = 400 = number of times the experiment is repeated.

As the sample size is quite large, i.e. <em>n</em> = 400 > 30 the central limit theorem can be used to approximate the sampling distribution of the sample mean.

The mean of the distribution of sample means is:

\mu_{\bar x}=\mu=3.5

The standard deviation of the distribution of sample mean is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{1.7078}{\sqrt{400}}=0.0854

The distribution of the sample mean is:

\bar X\sim N(\mu = 3.5,\ \sigma=0.0854).

Thus, the correct option is (C).

4 0
3 years ago
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