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fgiga [73]
3 years ago
8

I don’t know how to label the sides as “opposite, adjacent, hypotenuse” so I’m not sure which formula to use

Mathematics
1 answer:
ivanzaharov [21]3 years ago
3 0

Step-by-step explanation:

The opposite side is the side of the triangle facing the given angle. The adjacent side is the side that the given angle and the right angle lies on. The hypotenuse side is the side facing the right angle. (The hypotenuse side usually looks like a slanted line. Do note that this is not true if your adjacent and hypotenuse side is not horizontal and vertical respectively on the plane of your paper.)

I'll explain a bit more below using the some of the questions.

Q1) The given angle here is 27°.

X is facing the given angle hence it is the opposite side.

The side that is 10 units long is the adjacent side as 27° and the right angle sits on it.

tanθ= opp/adj

tan27°= x/ 10

x= 10(tan 27°)

x= 5.1 units (nearest tenth)

Q3) x is the adjacent side, while the side that is 7 units long is the hypotenuse side.

cosθ= adj/hyp

cos39°= x/7

x= 7(cos39°)

x= 5.4 units (nearest tenth)

Q4) x is the hypotenuse and the opposite side is 8 units long.

sinθ= opp/hyp

sin46°= 8/x

x(sin46°)= 8

x= 8/ sin46°

x= 11.1 units (nearest tenth)

Q5) x is facing the given angle, hence it is the opposite side. The adjacent side is given here, which is 6 units long.

tanθ= opp/adj

tan23°= x/6

x= 6(tan 23°)

x= 2.5 units (nearest tenth)

Feel free to ask if you still do not understand any concepts :)

sinθ= opp/hyp

cosθ= adj/hyp

tanθ= opp/adj

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Suppose that a random sample of 10 newborns had an average weight of 7.25 pounds and sample standard deviation of 2 pounds. a. T
frosja888 [35]

Answer:

z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565  

p_v =P(Z  

a) If we compare the p value and the significance level given \alpha=0.05 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 true average is not significantly less than 7.5.  

b) \chi^2 =\frac{10-1}{1.96} 4 =18.367  

p_v =P(\chi^2 >18.367)=0.0311

If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

Step-by-step explanation:

Assuming this info: "Suppose birth weights follow a normal distribution with mean 7.5 pounds and standard deviation 1.4 pounds"

1) Data given and notation  

\bar X=7.25 represent the sample mean  

s=1.2 represent the sample standard deviation

\sigma=1.4 represent the population standard deviation

n=10 sample size  

\mu_o =7.5 represent the value that we want to test  

\alpha=0.05,0.01 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

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

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is less than 7.5, the system of hypothesis would be:  

Null hypothesis:\mu \geq 7.5  

Alternative hypothesis:\mu < 7.5  

Since we know the population deviation, 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".  

3) Calculate the statistic  

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

z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565  

4)P-value  

Since is a left tailed test the p value would be:  

p_v =P(Z  

5) Conclusion  

Part a

If we compare the p value and the significance level given \alpha=0.05 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 true average is not significantly less than 7.5.  

Part b

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=10 represent the sample size

\alpha=0.01 represent the confidence level  

s^2 =4 represent the sample variance obtained

\sigma^2_0 =1.96 represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance increase, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 \leq 1.96

Alternative hypothesis: \sigma^2 >1.96

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{10-1}{1.96} 4 =18.367

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 9. And since is a right tailed test the p value would be given by:

p_v =P(\chi^2 >18.367)=0.0311

In order to find the p value we can use the following code in excel:

"=1-CHISQ.DIST(18.367,9,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

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Answer:

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Step-by-step explanation:

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cross multiply and you get

100/

divide both by 2 and you get 50/4 then you divide both by 2 again and you get 25/2 because you can't reduce it anymore

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