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Alborosie
2 years ago
11

There are about 4200 college presidents in the United States, and they have annual incomes with a distribution that is skewed in

stead of being normal. Many different samples of 40 college presidents are randomly selected, and the mean annual income is computed for each sample. What is the approximate shape of the sample means?
Mathematics
1 answer:
forsale [732]2 years ago
3 0
<span>Because a mean is an average, the means of each of the samples of 40 college presidents' incomes will have less of a skew than the actual income distribution in dollars of all of them. That is because outliers, like a president who is paid $2 million a year, will be averaged in with many others who are paid less. The shape of the curve of the means will depend upon how many different samples of 40 presidents are included, with the shape becoming more bell-like (normal distribution) the more random samples are included. Another consideration is how often the outliers are included and how big the income skew is that you start with. For example, if 2 college presidents are earning $2 million a year and the other 4,198 are earning $200,000 or less then the shape of your curve will depend greatly upon the number of times those two high earners are randomly selected and factored into the mean.</span>
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Answer:

Step-by-step explanation:

we know

\vec{i}\times \vec{j}=\vec{k}

\vec{j}\times \vec{k}=\vec{i}

\vec{k}\times \vec{i}=\vec{j}

(a)\left [ \left ( \hat{i}+\hat{j}\right )\times \hat{i}\right ]\times \hat{j}

=\left [ \hat{i}\times \hat{i}+\hat{j}\times \hat{i}\right ]\times \hat{j}

=\left [ 0-\hat{k}\right ]\times \hat{j}

=\hat{i}

(b)\left ( \hat{j}+\hat{k}\right )\times \left ( \hat{j}\times \hat{k}\right )

=\left ( \hat{j}+\hat{k}\right )\times \left ( \hat{i}\right )

=\hat{k}+\hat{j}

(c)4\hat{i}\times \left ( \hat{i}+\hat{j}\right )

=4\hat{i}\times \hat{i}+4\hat{i}\times \hat{j}

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(d)\left ( \hat{k}+\hat{j}\right )\times \left ( \hat{k}-\hat{j}\right )

=\hat{k}\times \hat{k}-\hat{k}\times \hat{j}+\hat{j}\times \hat{k}-\hat{j}\times \hat{j}

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2 years ago
Jason paints 1/4 of the area of his living room walls, w, on Monday. On Tuesday, he paints twice as much as he painted on Monday
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6 0
2 years ago
1:Suppose we have a bag with $10$ slips of paper in it. Eight slips have a $3$ on them and the other two have a $9$ on them. Wha
3241004551 [841]

Answer:

Step-by-step explanation:

1 ) No of total slips after addition = 11

8 slip with 3 on it

3 slips with 9  on it

expectation value = probability of 3 x 3 + probability of 9 x 9

=  (8 / 11)  x 3 + (3 / 11)  x 9

24 / 11 + 27 / 11 = 4.636

2 )

No of total slips after addition = 12

8 slip with 3 on it

4 slips with 9  on it

expectation value = probability of 3 x 3 + probability of 9 x 9

=  (8 / 12)  x 3 + (4 / 12)  x 9

2 + 3 = 5

3 )

Let n be the required number

No of total slips after addition = 10+n

8 slip with 3 on it

2 + n  slips with 9  on it

expectation value = probability of 3 x 3 + probability of 9 x 9

=  (8 / 10+n )  x 3 + (2+n  / 10+n )  x 9 = 6

24 + 18 + 9n / 10 + n  = 6

42 + 9n = 60 + 6n

3 n = 18

n = 6

4 )

Let n be the required number

No of total slips after addition = 10+n

8 slip with 3 on it

2 + n  slips with 9  on it

expectation value = probability of 3 x 3 + probability of 9 x 9

=  (8 / 10+n )  x 3 + (2+n  / 10+n )  x 9 = 8

24 + 18 + 9n / 10 + n  = 8

42 + 9n = 80 + 8n

n =

n = 38

Minimum of 38 has to be added .

6 0
2 years ago
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andrew-mc [135]
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So the answer is 144</span><span>°.</span>
8 0
2 years ago
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inysia [295]

Luckily for us, this problem solved the bulk of its own question!


Now, all we have to do is solve for "x", or the amount of fiction books and plug its value back into the original equation to find the value of "y", or the amount of nonfiction books.


2x = 38

x = 19

There are 19 fiction books.

x + y = 26

19 + y = 26

y = 7

There are 7 nonfiction books.

Hope this helps!

7 0
2 years ago
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