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3 years ago

Find the prime factorization of each number. 1. 50 =

Mathematics

1 answer:

Varvara68 [4.7K]3 years ago

7 0

Answer:

1.5 is not a prime number

Step-by-step explanation:

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For a sample of n=64, the probability of a sample mean being less than 20.5 if u = 21 and sigma = 1.31 is (Round to four decim

Svetlanka [38]

Answer:

The mean is usual

Step-by-step explanation:

We have that the mean (m) is equal to 21, the standard deviation (sd) = 1.31 and the sample size (n) = 64

They ask us for P (x <20.5)

For this, the first thing is to calculate z, which is given by the following equation:

z = (x - m) / (sd / (n ^ 1/2))

We have all these values, replacing we have:

z = (20.5 - 21) / (1.31 / (64 ^ 1/2))

z = -3.05

With the normal distribution table (attached), we have that at that value, the probability is:

P (z <-3.05) = 0.0002

The mean is usual because P (x> 20.5) = 1 - P (x <20.5) = 1 - 0.0002 = 0.9998 is a fairly high probability.

6 0

3 years ago

Can i get the answer please

Sladkaya [172]

No, because 40 x 10 is not 4000, but is is 400. It is 100x greater not 10x

8 0

3 years ago

.........................................................

Studentka2010 [4]

Answer: no cheating on the state test

Step-by-step explanation:

3 0

2 years ago

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Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage ea

garik1379 [7]

Answer:

We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.

Step-by-step explanation:

We have a sample of executives, of size n=160, and the proportion that prefer trucks is 26%.

We have to calculate a 95% confidence interval for the proportion.

The sample proportion is p=0.26.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.26*0.74}{160}}\\\\\\ \sigma_p=\sqrt{0.0012}=0.0347$

The critical z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=1.96 \cdot 0.0347=0.068$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.26-0.068=0.192\\\\UL=p+z \cdot \sigma_p = 0.26+0.068=0.328$

The 95% confidence interval for the population proportion is (0.192, 0.328).

We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.

3 0

3 years ago

Whole numbers greater than 10 but less than 20

hichkok12 [17]

11, 12, 13, 14, 15, 16, 17, 18, 19 Sorry if it doesn't help, I'm being too obvious

4 0

3 years ago

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