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ludmilkaskok [199]
3 years ago
8

Someone please help. Will mark brainliest!

Mathematics
2 answers:
mars1129 [50]3 years ago
7 0

Hello!

I believe the answer A) Step 2, replace perpendicular bisector with angle bisector.

I hope it helps!

marusya05 [52]3 years ago
5 0

Step 2, replace the perpendicular bisector with angle bisector.

Note that in Step 3, it says "bisectors" with an s. This means that there are more than one angle bisector. Step 1 only states 1 bisector, so above answer should be correct.

hope this helps

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The difference between the most hours of sleep and the least of these students was 2.5 hours. True or false?​
SashulF [63]

The correct anwer is False

Explanation

According to the graph, it can be seen that a student has four hours of sleep as the minimum number of hours of sleep; two students have six hours of sleep, four students have six and a half hours of sleep, four have seven hours of sleep, three have seven and a half hours of sleep, five have eight hours of sleep, and one has eight and a half hours of sleep as maximum hours of sleep. Therefore, it can be affirmed that the statement that the difference between the maximum amount and the minimum number of hours is two and a half hours is false because between four hours and eight and a half hours there are four and a half hours of difference. So, the correct answer is False.

8 0
3 years ago
Please answer asap this is 7th grade math<br> find the area
Rainbow [258]

25pi

pi x 10^2 is 100pi

Divide by four = 25pi

7 0
2 years ago
Read 2 more answers
Suppose that a random sample of 10 newborns had an average weight of 7.25 pounds and sample standard deviation of 2 pounds. a. T
frosja888 [35]

Answer:

z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565  

p_v =P(Z  

a) If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.  

b) \chi^2 =\frac{10-1}{1.96} 4 =18.367  

p_v =P(\chi^2 >18.367)=0.0311

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 higher than 1.96.

Step-by-step explanation:

Assuming this info: "Suppose birth weights follow a normal distribution with mean 7.5 pounds and standard deviation 1.4 pounds"

1) Data given and notation  

\bar X=7.25 represent the sample mean  

s=1.2 represent the sample standard deviation

\sigma=1.4 represent the population standard deviation

n=10 sample size  

\mu_o =7.5 represent the value that we want to test  

\alpha=0.05,0.01 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is less than 7.5, the system of hypothesis would be:  

Null hypothesis:\mu \geq 7.5  

Alternative hypothesis:\mu < 7.5  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

3) Calculate the statistic  

We can replace in formula (1) the info given like this:  

z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565  

4)P-value  

Since is a left tailed test the p value would be:  

p_v =P(Z  

5) Conclusion  

Part a

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.  

Part b

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=10 represent the sample size

\alpha=0.01 represent the confidence level  

s^2 =4 represent the sample variance obtained

\sigma^2_0 =1.96 represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance increase, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 \leq 1.96

Alternative hypothesis: \sigma^2 >1.96

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{10-1}{1.96} 4 =18.367

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 9. And since is a right tailed test the p value would be given by:

p_v =P(\chi^2 >18.367)=0.0311

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

"=1-CHISQ.DIST(18.367,9,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 higher than 1.96.

4 0
3 years ago
The measure of the angle
miv72 [106K]
First, we know this diagram consists of two horizontal lines cut by a transversal line. Therefore, we know that the given angle that measures 113° and the angle we want to find are alternate interior angles. Since all alternate interior angles are equal, we know the unknown angle must also be 113°.

I hope this helps.
5 0
3 years ago
Read 2 more answers
The vertices of square WXYZ are W(-2, -1), X(-2,-5),Y(-6,-5), and Z(-6,-1) what is the perimeter and area of the square show you
labwork [276]
Line Segment WX=4
Line Segment XY=4 
Line Segment YZ=4
Line Segment ZW=4
4+4+4+4=16
The perimeter is 16 units.
7 0
3 years ago
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