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Alecsey [184]
2 years ago
12

According to the National Postsecondary Student Aid Study conducted by the U.S. Department of Education in 2008, 62% of graduate

s from public universities had student loans.
We randomly select 50 students at a time.
What is the standard error for the distribution of sampling proportions? If necessary, round to three decimal places.
Mathematics
1 answer:
Ronch [10]2 years ago
6 0

The standard error for the distribution of sampling proportions will be 0.0686.

<h3>What is standard error for the distribution of sampling proportions?</h3>

In mathematics, the difference between a data set and the populace's true average is known as primary data deviation from the mean.

According to the National Postsecondary Student Aid Study conducted by the U.S.

Department of Education in 2008, 62% of graduates from public universities had student loans.

We randomly select 50 students at a time.

Then the standard error for the distribution of sampling proportions will be

\rm Standard\  error = \sqrt{\dfrac{p(1-p)}{n}}\\\\Standard\  error = \sqrt{\dfrac{0.62(1-0.62)}{50}}\\\\Standard\  error = 0.0686

More about the standard error for the distribution link is given below.

brainly.com/question/14524236

#SPJ1

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So lets get to the problem 

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<span>To make it easier I'm going to write the same thing like this </span>
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<span>Cos165° </span>
<span>= Cos ( 90° + 45°+30° ) </span>
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<span>= Sin45°Cos30° + Cos45°Sin30° </span>


<span>Tan165° </span>
<span>= Tan ( 90° + 45°+30° ) </span>
<span>= -Cot( 45°+30° )..... (∵Cot(90 + θ)=-Tanθ </span>
<span>= -1/tan(45°+30°) </span>
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<span>Substitute the above values with the following... These should be memorized </span>
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Answer: y = -\frac{1}{6}x-5

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Answer:

Step-by-step explanation:

Part A.

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