I need help like really bad this is due today

If a 45 degree 45 degree 90 degree triangle has a hypotenuse length of 12 find the leg length

AURORKA [14]

Answer:

<h2>Leg length = 8.5</h2><h2 />

Step-by-step explanation:

let x represent the leg length

if we use the Pythagorean theorem we get:

x² + x² = 12²

then

2x² = 12²

Then

x² = 144/2 = 72

then

x = √(72)

=8,485281374239

7 0

2 years ago

A college infirmary conducted an experiment to determine the degree of relief provided by three cough remedies. Each cough remed

KiRa [710]

Answer:

$p_v = P(\chi^2_{4,0.05} >3.81)=0.43233$

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed. So we can say that the 3 remedies ar approximately equally effective.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

NyQuil Robitussin Triaminic Total

No relief 11 13 9 33

Some relief 32 28 27 87

Total relief 7 9 14 30

Total 50 50 50 150

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference in the three remedies

H1: There is a difference in the three remedies

The level os significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{50*33}{150}=11$

$E_{2} =\frac{50*33}{150}=11$

$E_{3} =\frac{50*33}{150}=11$

$E_{4} =\frac{50*87}{150}=29$

$E_{5} =\frac{50*87}{150}=29$

$E_{6} =\frac{50*87}{150}=29$

$E_{7} =\frac{50*30}{150}=10$

$E_{8} =\frac{50*30}{150}=10$

$E_{9} =\frac{50*30}{150}=10$

And the expected values are given by:

NyQuil Robitussin Triaminic Total

No relief 11 11 11 33

Some relief 29 29 29 87

Total relief 10 10 10 30

Total 50 50 50 150

And now we can calculate the statistic:

$\chi^2 = \frac{(11-11)^2}{11}+\frac{(13-11)^2}{11}+\frac{(9-11)^2}{11}+\frac{(32-29)^2}{29}+\frac{(28-29)^2}{29}+\frac{(27-29)^2}{29}+\frac{(7-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(14-10)^2}{10} =3.81$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(3-1)=4$

And we can calculate the p value given by:

$p_v = P(\chi^2_{4,0.05} >3.81)=0.43233$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.81,4,TRUE)"

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed.

4 0

2 years ago

A business school has a goal that the average number of years of work experience of its MBA applicants is more than three years.

gizmo_the_mogwai [7]

Answer:

$z=\frac{3.1-3}{\frac{2.449}{\sqrt{47}}}=0.280$

$p_v =P(z>0.280)=0.390$

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 reject the null hypothesis, so we can't conclude that the height of men actually its NOT significant higher than 0.3 at 1% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=3.1$ represent the sample mean

$\sigma=\sqrt{6}$ represent the sample standard deviation for the sample

$n=47$ sample size

$\mu_o =3$ 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 3, the system of hypothesis would be:

Null hypothesis: $\mu \leq 3$

Alternative hypothesis: $\mu > 3$

If we analyze the size for the sample is > 30 and we know the population deviation so 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".

Calculate the statistic

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

$z=\frac{3.1-3}{\frac{2.449}{\sqrt{47}}}=0.280$

P-value

Since is a one side test the p value would be:

$p_v =P(z>0.280)=0.390$

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 reject the null hypothesis, so we can't conclude that the height of men actually its NOT significant higher than 0.3 at 1% of signficance.

3 0

2 years ago

For every penny Sam puts into his bank, Tara puts 4 pennies into her bank. If Sam puts 10 pennies into his bank, how many pennie

devlian [24]

Answer: If Tara puts 4 pennies in for every one Sam does then after he puts in 10 she would of put 40.

She put in 40 pennies. Hope this helps ;)

3 0

2 years ago

NEED HELP 45 POINTS PLZ

miskamm [114]

Answer:

I have given the answer in the chat box as i was not able to post officialy

Step-by-step explanation:

6 0

2 years ago