Please help me I have 2 questions.

andre [41]

Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis.

The domain is the values of x between 1 and 7

The range

4 0

1 year ago

I need help with this <br> ASAP!!

liraira [26]

Answer:=

−243 /512 Fraction

4 0

3 years ago

Which of the following is a step in simplifying the expression x multiplied by y to the power of 3 over x to the power of negati

stich3 [128]

Answer:

-3

Step-by-step explanation:

5 0

3 years ago

given sin A =3/-/34 and that angle S is in quadrant 1, find the exact value of tan A in simplest radical form using a ration

Vedmedyk [2.9K]

Answer:

$\tan(A) = \frac{3}{5}$

Step-by-step explanation:

Given

$\sin(A) = \frac{3}{\sqrt {34}}$

$0 \le A \le 90$ --- First Quadrant

Required

Find tan(A)

The sin of an angle is:

$\tan(A) = \frac{Opposite}{Hypotenuse}$

and

$\sin(A) = \frac{3}{\sqrt {34}}$

By comparison:

$Opposite = 3$

$Hypotenuse = \sqrt{34$

So, the Adjacent is:

$Hypotenuse^2 = Adjacent^2 + Opposite^2$

$(\sqrt{34})^2 = Adjacent^2 + 3^2$

$34 = Adjacent^2 + 9$

Collect like terms

$Adjacent^2 =34 - 9$

$Adjacent^2 =25$

Take square roots

$Adjacent =\sqrt{25$

$Adjacent =5$

The tangent of an angle is:

$\tan(A) = \frac{Opposite}{Adjacent}$

This gives:

$\tan(A) = \frac{3}{5}$

8 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

3 years ago