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damaskus [11]
3 years ago
8

What's the surface area of a sphere with radius 5 centimeters

Mathematics
1 answer:
torisob [31]3 years ago
3 0
Surface area: 4*pi*r^2 = 4*pi*5^2= 4*25*pi = 100pi or <span>314.16</span>
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A fair six-sided die is rolled 72 times. Theoretically, about how many times should a roll of "4" occur?
Fynjy0 [20]

Answer:

OB. 12

Step-by-step explanation:

Since the die is a fair six-sided die and the question uses should... we know that a roll of "4" has a 1/6 chance to land every time the die is rolled.

1. 72 rolls * 1/6 chance to roll a 4 = 12 times

7 0
3 years ago
HOLA necesito ayuda... como puedo combertir 50% en fraccion y decimal. Y como combertir 25% en fraccion y decimal
stepladder [879]

Answer:

Step-by-step explanation:

50/100 = 1/2 or 0.5

25/100 = 1/4 or 0.25

8 0
2 years ago
A patient tells the nurse that she has smoked two packs of cigarettes a day for 20 years. the nurse records this as how many pac
Zina [86]
There are 365 days in a year 
2 packs x 365 days = 730 packs per year

check work:

730 divided by 2 = 365

5 0
4 years ago
PLS HELP ME
lutik1710 [3]

Answer: The correct option is 2.

4 0
2 years ago
Read 2 more answers
Suppose a mutual fund qualifies as having moderate risk if the standard deviation of its monthly rate of return is less than 5​%
Leokris [45]

Answer:

\chi^2 =\frac{28-1}{25} 21.7156 =23.453

p_v =P(\chi^2

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(23.453,27,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly lower than 5% so we can't conclude that we have a moderate risk for this case.

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

n=28 represent the sample size

\alpha=0.1 represent the confidence level  

s^2 =21.7156 represent the sample variance obtained

\sigma^2_0 =25 represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance specification is lower than 25, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 \geq 25

Alternative hypothesis: \sigma^2

Calculate the statistic  

For this test we can use the following statistic:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

\chi^2 =\frac{28-1}{25} 21.7156 =23.453

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 27. And since is a right tailed test the p value would be given by:

p_v =P(\chi^2

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(23.453,27,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly lower than 5% so we can't conclude that we have a moderate risk for this case.

4 0
3 years ago
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