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Nikitich [7]
3 years ago
7

Wich are factors of 31

Mathematics
2 answers:
RideAnS [48]3 years ago
6 0

31 is prime, so the only factors are 31 and 1

disa [49]3 years ago
3 0

1 and 31

Hope this helps

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A frozen dinner at the grocery store is purchased for $2.20 a wholesale and is a marked up for 40 percent. a customer wants to b
iren [92.7K]

First of all, you surely do have a lot of incorrect grammar in your statement. I can't understand a word that you are asking.

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The annual interest rate of Mike's savings account is 4.8%, and simple interest is calculated monthly. What is the periodic inte
Reptile [31]

The correct option will be: d. 0.4%

For finding the periodic interest rate , we need to divide the annual interest rate by the number of times interest calculated in a year.

Here, the simple interest is calculated monthly, that means the number of times interest calculated in a year will be 12.

Annual interest rate is 4.8%

So, the periodic interest rate = \frac{4.8}{12}= 0.4%

5 0
3 years ago
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One leg of a right triangle is 28 inches longer than the other leg, and the hypotenuse is 52 inches. Find the lengths of the leg
katovenus [111]

Step-by-step explanation:

Let x represent the missing sides measurement.

x+x+28=52

combine like terms

2x + 28 = 52

subtract 28 from both sides

2x + 28 = 52

     - 28  - 28

2x = 24

divide both sides by 2

x = 12

plug it back in

one leg is 12 inches

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7 0
3 years ago
A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
valentina_108 [34]

Answer:

(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b) The value of P(\bar X is 0.7642.

(c) The value of P(\bar X\geq 69.1) is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

\mu=68\\\sigma=15

(a)

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

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

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

\mu_{\bar x}=\mu=68

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

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X is 0.7642.

(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

                    =P(\bar X>69.6)

                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

Thus, the value of P(\bar X\geq 69.1) is 0.3670.

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