Help me please and thnk you

Find the area of the shaded region.

densk [106]

[1]

A1 = (h (a + b)) / 2

A1 = (21 (17 + 32)) / 2

A1 = (21 x 49) / 2

A1 = 1,029 / 2

A1 = 514.5 mm²

[2]

A2 = (b x h) / 2

A2 = (11 x 9) / 2

A2 = 99 / 2

A2 = 49.5 mm²

[3]

The area of the shaded region =

A1 - A2 =

514.5 mm² - 49.5 mm² =

465 mm²

The answer is 465 mm².

4 0

3 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

the diagonal of a rectangular picture frame is 12 inches and it's length is 9 inches what is the width of the picture frame ?

sladkih [1.3K]

I think it would be 5

4 0

3 years ago

What dose No mean in math

Wewaii [24]

Answer:

Nobelium

Step-by-step explanation:

That looks like a chemistry element unless No is meaning the abbreviation No. - Meaning, number.

7 0

3 years ago

On a map the distance between rip and Janeiro Brazil and San Jose Costa Rica is 4 inches.the scale on the map shows that 1 inch=

Vera_Pavlovna [14]

Answer:

The way a proportion is set is as follows;

$\dfrac{1 \ inch}{200 \ miles } = \dfrac{4 \ inches}{ Distance \ from \ Rio \ de \ Janeiro \ to \ San \ Jose}$

The Distance from Rio de Janeiro and San Jose Costa Rica is then found as 800 miles

Step-by-step explanation:

The dimension of the distance between Rio de Janeiro and San Jose Costa Rica = 4 inches

The map scale = 1 inch to 200 miles

To figure out how many miles it is or the actual distance from Rio de Janeiro and San Jose Costa Rica, we have;

$\dfrac{1 \ inch}{200 \ miles } = \dfrac{4 \ inches}{ Distance \ from \ Rio \ de \ Janeiro \ to \ San \ Jose}$

Therefore, we have;

(4 inches × 200 miles)/(1 inche) = Distance from Rio de Janeiro and San Jose Costa Rica

Which gives;

Distance from Rio de Janeiro and San Jose Costa Rica = 800 miles.

4 0

3 years ago