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yarga [219]
3 years ago
11

THE FUNCTION ARCSINE CAN ALSO BE DEFINED AS?

Mathematics
1 answer:
vodomira [7]3 years ago
3 0
The arcsine, \sin^{-1}, is the inverse of the \sin function. This means that it takes as <em>inputs </em>what would usually be <em>outputs </em>for the \sin function and produces as <em>outputs </em>what would usually be <em>inputs </em>for the \sin function.

This can be particularly useful when you're trying to find an angle on a right triangle, but you've only been given the lengths of the sides. To find any angle \theta in a right triangle, just take \sin^{-1} \frac{o}{h}, where o is the side opposite \theta and h is the hypotenuse of the right triangle.

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3537.73 cubic in

Step-by-step explanation:

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Diameter of the cone = 26 in

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Height of the cone = 20 in

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<u>Approx. 3537 . 73 cubic in (Ans)</u>

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3 0
3 years ago
The Scholastic Aptitude Test (SAT) is a standardized test for college admissions in the U.S. Scores on the SAT can range from 60
kodGreya [7K]

Answer:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

Step-by-step explanation:

Notation and previous concepts

n_1 represent the sample after the preparation

n_2 represent the sample without preparation  

\bar x_1 =678 represent the mean sample after preparation

\bar x_2 =1837 represent the mean sample without preparation

s_1 =197 represent the sample deviation after preparation

s_2 =328 represent the sample deviation without preparation

\alpha=0.1 represent the significance level

Confidence =90% or 0.90

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{(\frac{s^2_1}{n_s}+\frac{s^2_2}{n_s})} (1)  

The point of estimate for \mu_1 -\mu_2

The appropiate degrees of freedom are df=n_1+ n_2 -2

Since the Confidence is 0.90 or 90%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,df)  

The standard error is given by the following formula:  

SE=\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}  

After replace in the formula for the confidence interval we got this:

3.0 < \mu_1 -\mu_2

And we need to interpret this result:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

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3 years ago
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