Write the expression in factored form. X^2+3x+xy

The number of calories burned while jogging varies directly with the number of minutes spent jogging.

seropon [69]

Answer:

350.5

Step-by-step explanation:

3 0

3 years ago

Can some one help me with geometry

Anuta_ua [19.1K]

Um, what about geometry? What problem?

7 0

3 years ago

Marty has saved $72. He spent $8 on a video rental. Write a ratio as a fraction in simplest form to represent what portion of hi

Natalka [10]

The solution to the problem is as follows:

The savings left is just $72 - $8 = $64.
Therefore, the ratio is $64 : $72 = 8 : 9.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

3 0

4 years ago

Read 2 more answers

An automotive part must be machined to close tolerances to be acceptable to customers. Production specifications call for a maxi

Vilka [71]

Answer:

H0: $\sigma \leq 0.0004$

H1: $\sigma >0.0004$

$t=(30-1) [\frac{0.0224}{0.02}]^2 =36.25$

For this case since we have a right tailed test the p value is given by:

$p_v = P(\Chi^2_{29}>36.25)=0.166$

If we compare the p value and the significance level given we see that $p_v >\alpha$ so then we have enough evidence to FAIL to reject the null hypothesis that the true population variance it's lower or equal than 0.0004 at 5% of significance.

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

n = 30 sample size

$s^2 =0.0005$ represent the sample variance

s= 0.0224 represent the sample deviation

$\sigma_o =0.095$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic

$\alpha=0.05$ significance level

State the null and alternative hypothesis

On this case we want to check if the population variance is higher than 0.0004 (the specification), so the system of hypothesis are:

H0: $\sigma \leq 0.0004$

H1: $\sigma >0.0004$

In order to check the hypothesis we need to calculate th statistic given by the following formula:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

$t=(30-1) [\frac{0.0224}{0.02}]^2 =36.25$

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with df=30-1=19 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,29)". And our critical value would be $\Chi^2 =42.56$

Since our calculated value is lower than the critical value we to reject the null hypothesis.

What is the approximate p-value of the test?

For this case since we have a right tailed test the p value is given by:

$p_v = P(\Chi^2_{29}>36.25)=0.166$

If we compare the p value and the significance level given we see that $p_v >\alpha$ so then we have enough evidence to FAIL to reject the null hypothesis that the true population variance it's lower or equal than 0.0004 at 5% of significance.

6 0

3 years ago

The area covered by a spotlight on the side of a house is 954 square feet, as seen in the diagram below.

miss Akunina [59]

Answer:

$27.25\ ft^2$

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

we know that

The area on the side of the house that is NOT covered by the light is equal to the area of semicircle minus the area covered by a spotlight on the side of a house

Find the area of the semicircle

The area of a semicircle is equal to

$A=\frac{1}{2}\pi r^{2}$