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PIT_PIT [208]
3 years ago
6

Which of the following represents the prime factorization of 72?

Mathematics
2 answers:
Free_Kalibri [48]3 years ago
7 0

Answer:

2^3.3^2

Step-by-step explanation:

{2}^{3 ?}  = 8 \\  {?3}^{2?}  = 9 \\ 8 \times 9 = 72

Rasek [7]3 years ago
7 0

Answer:

2.2.2.3.3.

72 / 2 = 36

36/2 = 18

18/2 = 9

9/ 3= 3.

So the answer to your question is 2.2.2.3.3.

Hope this helps!

Have a great day and stay safe! :))))))

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A) multiply equation A by -2 so you get -2x - 2z = -12 and then you can add both equations to eliminate z
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The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 21 people reveals the mean yearly consum
PIT_PIT [208]

Correct question is;

The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 21 people reveals the mean yearly consumption to be 74 gallons with a standard deviation of 16 gallons.

a. What is the value of the population mean? What is the best estimate of this value?

b. Explain why we need to use the t distribution. What assumption do you need to make?

c. For a 90 percent confidence interval, what is the value of t?

d. Develop the 90 percent confidence interval for the population mean.

e. Would it be reasonable to conclude that the population mean is 68 gallons?

Answer:

A) Best estimate = 74 gallons

B) because the population standard deviation is unknown. The assumption we will make is that the population follows the normal distribution.

C) t = 1.725

D) 90% confidence interval for the population mean is (67.9772, 80.0228) gallons

E) Yes

Step-by-step explanation:

We are given;

Sample mean; x' = 74

Sample population; n = 21

Yearly Standard deviation; s = 16

A) We are not given the population mean.

So the closest estimate to the population mean would be the sample mean which is 74.

B) We are not given the population standard deviation and as such we can't use normal distribution. So what is used when population standard deviation is not known is called t - distribution table. The assumption we will make is that the population follows the normal distribution.

C) At confidence interval of 90% and DF = n - 1 = 21 - 1 = 20

From t-tables, the t = 1.725

D) Formula for the confidence interval is;

x' ± t(s/√n) = 74 ± 1.725(16/√21) = 74 ± 6.0228 = 67.9772 or 80.0228

Thus 90% confidence interval for the population mean is (67.9772, 80.0228) gallons

E) 68 gallons lies within the range of the confidence interval, thus we can say that "Yes, it is reasonable"

8 0
3 years ago
Can i please have some help??
Sliva [168]

Answer:

No solution

Step-by-step explanation:

When lines are parallel, they do not intersect and they do not have a solution.

Thank You

Hope it helps u

3 0
3 years ago
Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime i
OLEGan [10]
<h2>Answer with explanation:</h2>

Let \mu be the population mean lifetime of circulated $1 bills.

By considering the given information , we have :-

H_0:\mu=18\\\\H_a:\mu\neq18

Since the alternative hypotheses is two tailed so the test is a two tailed test.

We assume that the lifetime of circulated $1 bills is normally distributed.

Given : Sample size :  n=50 , which is greater than 30 .

It means the sample is large so we use z-test.

Sample mean : \overline{x}=18.8

Standard deviation : \sigma=2.8

Test statistic for population mean :-

z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

\Rightarrow\ z=\dfrac{18.8-18}{\dfrac{2.8}{\sqrt{50}}}\approx2.02

The p-value= 2P(z>2.02)=0.0433834

Since the p-value (0.0433834) is greater than the significance level (0.02) , so we do not reject the null hypothesis.

Hence, we conclude that we do not have enough evidence to support the alternative hypothesis that the average lifetime of a circulated $1 bill differs from 18 months.

6 0
3 years ago
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