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Alisiya [41]
3 years ago
11

Find the missing factor. 2n^2(n - 3) - 4(n - 3) = ( ) (n - 3)

Mathematics
1 answer:
Molodets [167]3 years ago
8 0

<em>Hey</em><em>!</em><em>!</em><em>!</em><em>!</em>

<em>Here</em><em>'s</em><em> </em><em>your</em><em> </em><em>answer</em>

<em>The</em><em> </em><em>missin</em><em>g</em><em> </em><em>factor</em><em> </em><em>is</em><em> </em><em>(</em><em>2</em><em>n</em><em>^</em><em>2</em><em>-</em><em>4</em><em>)</em>

<em>Hope </em><em>it</em><em> </em><em>helps</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>

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I need help in problem 8 plz help me I need a a
iogann1982 [59]
Here are three ways where There could be equal rows

3*12=36
4*9=36
6*6=36
6 0
2 years ago
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
What is the slope of a line perpendicular to the line passing through (1, -3) and (-4, 7)?
andrezito [222]

Answer:

-2

Step-by-step explanation:

m=\frac{y2-y1}{x2-x1}

7 0
3 years ago
What is the negative reciprocal of 2
Zinaida [17]

Answer:

- \frac{1}{2}

Step-by-step explanation:

The reciprocal of a number n is \frac{1}{n}

The negative reciprocal is the negative of this, thus

reciprocal of 2 is - \frac{1}{2}

3 0
3 years ago
PLEASEEEEE HELLPPPP ME I NEED HELP ASAPPP I CANNOT FAIL THIS!!!!!!!
Elodia [21]

Answer:

1. Both Dan and Tess are correct.

    Tess:

         0.15 * 24.50 = 3.675

         24.5 - 3.675 = 20.825

         0.1 * 20.825 = 2.0825

         20.825 + 2.0825 = 22.9075

    Dan:

         0.85 * 24.50 =  20.825

         20.825 * 1.1 = 22.9075

    Both Dan and Tess used methods that provided the correct answer. Both methods follow the same path, but Dan's method simply took less steps since he used percentages that would take out the correct number, unlike Tess' percentages which would give you the number that you would then need to take out.

2. First, convert Amy's and Mike's numbers to a decimal of the whole so it is easier to compare the numbers.

    13% = 0.13 of the whole

    Dan has 0.14 of the whole

    0.5/6 = 0.083.. of the whole

Dan used the greatest amount of milk, using 0.14 of the 6 gallon container.

3.

    2p + 12 = 58 | Create an equation to show the prices relations

    2p = 46 | Take out the price of the concert ticket to leave us with the passes.

    p = 23 | Divide by 2 to find the cost of each pass.

4.

    x ≤ (18 - 6.5 - 5.75)/1.15

    18 - 6.50 = 11.5 | Take out the price of the notebook from the total.

    11.5 - 5.75 = 5.75 | Take out the amount of money that she wishes to save.

    5.75 / 1.15 = 5 | Divide by the price of each candy to find the max amount that she can buy.

Andrea has enough money to buy up to 5 candy bars.

x ≤ 5

8 0
3 years ago
Read 2 more answers
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