What is equifenlnt to 35000 cm in kelomiters

To figure this out, always divide y by x in each table. All of the values have to be equal or the table isn’t proportional.

In this case, the last choice is proportional.

7 0

3 years ago

What is the measure of angle Q to the nearest whole degree?<br> 43°<br> 49°<br> 53°<br> 58°

geniusboy [140]

Answer:

m∠Q ≈ 53°

Step-by-step explanation:

To find the measure of ∠Q, the law of cosines will need to be used. Lowercase letters represent the side lengths, while upper case letters represent angles.

In this situation, 'A' will be ∠Q. Therefore:

17² = 18² + 20² -2(18)(20)cosQ

Simplify:

289 = 324 + 400 -2(360)cosQ

Continue simplifying down:

-435 = -720cosQ

Divide both sides by '-720':

0.604 = cosQ

$cos^{-1}Q = 0.604$

m∠Q ≈ 52.83 or 53° rounded to the nearest whole degree.

3 0

3 years ago

Suppose a local manufacturing company claims their production line has a variance of less than 9.0. A quality control engineer d

Alexxx [7]

Answer:

$\chi^2 =\frac{35-1}{9} 2.12^2 =16.979$

$p_v =P(\chi^2$

If we compare the p value and the significance level provided we see that $p_v$ so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And that means that the population variance is significantly lower than 9

Step-by-step explanation:

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"

$n=35$ represent the sample size

$\alpha=0.01$ represent the confidence level

$s =2.12$ represent the sample deviation obtained

$\sigma_0 =3$ 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 9 and the deviation lower than 3, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 9$

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.

$\chi^2 =\frac{35-1}{9} 2.12^2 =16.979$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case df= n-1= 35-1=34. And since is a left 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(16.979,34,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that $p_v$ so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And that means that the population variance is significantly lower than 9

6 0

4 years ago