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3 years ago

Find cscθ.pls answer

Mathematics

1 answer:

ExtremeBDS [4]3 years ago

7 0

Answer:

The answer to your question is csc Ф = $\frac{-5}{4}$

Step-by-step explanation:

Process

1.- Determine the sign

We must determine csc Ф in the forth quadrangle, here csc is negative.

2.- Determine the hypotenuse

c² = a² + b²

c² = 3² + (-4)²

c² = 9 + 16

c² = 25

c = 5

3.- Determine csc Ф

csc Ф = $\frac{hypotenuse}{opposite side}$

csc Ф = $\frac{5}{-4}$ = $\frac{-5}{4}$

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a) For this case we can conclude that we are 95% confident that the true mean for the variable of interest is between 3.4 and 4.24 days in the US population.

b) The 95% confident means that if we select 100 different samples and calculate the 95% confidence interval for each sample selected, then we will have approximately 95 out of 100 confidence intervals will contain the true mean for the parameter of interest.

c) If we have the same info but we want more confidence that implies that the confidence interval would be wider, since the margin of error increase with more confidence.

d) Assuming that we have the same confidence level and the value for the deviation not changes. If we see if we decrease the sample size, then the margin of error would be lower since the original sample size was 1151. So then if we use 500 Americans we would have a lower value for the margin of error for this new interval. And then our confidence interval would smaller than the original.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The confidence interval after apply the formulas was (3.40 ,4.24)

Part a

For this case we can conclude that we are 95% confident that the true mean for the variable of interest is between 3.4 and 4.24 days in the US population.

Part b

The 95% confident means that if we select 100 different samples and calculate the 95% confidence interval for each sample selected, then we will have approximately 95 out of 100 confidence intervals will contain the true mean for the parameter of interest.

Part c

If we have the same info but we want more confidence that implies that the confidence interval would be wider, since the margin of erroe increase with more confidence, because the critical value increase.

Part d

We need to take in count that the margin of error is given by:

$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$

Assuming that we have the same confidence level and the value for the deviation s not changes. If we see if we decrease the sample size, then the margin of error would be lower since the original sample size was 1151. So then if we use 500 Americans we would have a lower value for the margin of error for this new interval. And then our confidence interval would smaller than the original.

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3 years ago