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Katyanochek1 [597]

Answer:

H

Step-by-step explanation:

7 0

3 years ago

Solve m=<img src="https://tex.z-dn.net/?f=%5Cfrac%7By_%7B2%7D-%20%7By_%7B1%7D%7D%20%7D%7Bx_2-x_%7B1%7D%20%7D" id="TexFormula1" t

miss Akunina [59]

Answer:

$y_2 = (x_2 - x_1) +y_1$ .

Implied: $x_2\ne x_1$ .

Step-by-step explanation:

Get rid of the denominator by multiplying both sides of this equation with $(x_2 - x_1)$ :

$m(x_2 - x_1) = y_2 - y_1$ .

Isolate $y_2$ by moving $(-y_1)$ to the left-hand side of the equation. That is: add $y_1$ to both sides of this equation.

$m(x_2 - x_1) + y_1 = y_2$ .

By the symmetric property of equalities,

$y_2 = m(x_2 - x_1) + y_1$

6 0

3 years ago

A university claims that the average cost of books per student, per semester is $300. A group of students believes that the actu

LenKa [72]

Answer:

$t=\frac{345-300}{\frac{200}{\sqrt{100}}}=2.25$

$p_v =P(t_{(99)}>2.25)=0.0133$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.01$ 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 mean is higher than 300 at 1% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=345$ represent the sample mean

$s=200$ represent the sample standard deviation

$n=100$ sample size

$\mu_o =300$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 300, the system of hypothesis would be:

Null hypothesis: $\mu \leq 300$

Alternative hypothesis: $\mu > 300$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{345-300}{\frac{200}{\sqrt{100}}}=2.25$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=100-1=99$

Since is a one right tailed test the p value would be:

$p_v =P(t_{(99)}>2.25)=0.0133$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.01$ 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 mean is higher than 300 at 1% of significance.

7 0

4 years ago

What is the solutionif this is your equation 9+k=18

icang [17]

Answer:

9k=18

k=9

Step-by-step explanation:

7 0

3 years ago

What is 2 divided by 1054 in long division

Trava [24]