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pochemuha
3 years ago
8

Describe the sampling distribution of Phat. Assume the size of the Population is 25,000. N = 500 P = 0.4 Determine the Mean of t

he sampling distribution of Phat. Determine the standard deviation of the sampling distribution of Phat.

Mathematics
1 answer:
PIT_PIT [208]3 years ago
5 0

Answer:

0.022

Step-by-step explanation:

Given that :

Population size = 25000

n = 500 ; p = 0.4

Size of random sample (n) = 500

5% of population size : 0.05 * 25000 = 1250

Distribution is normally distributed since n < 5% of population size

Hence, the mean of the distribution = p = 0.4

Standard deviation = √((pq) /n)

q = 1 - p ; q = 1 - 0. 4 = 0.6

Standard deviation = √((0.4 * 0.6) /500)

Standard deviation = 0.0219089

= 0.022

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The point A(–1, 4) is translated five units right and three units down. Which rule describes the translation?
lozanna [386]

Answer: SECOND OPTION.

Step-by-step explanation:

Given the following point identified as "A":

A(-1,4)

You can identify that the x-coordinate of the point is:

x=-1

And its y-coordinate is:

y=4

According to the exercise, this point is translated five units right and three units down. This means that, in order to find the new coordinates, you need to add 5 to the original x-coordinate and subtract 3 from the original y-coordinate.

Therefore, you can conclude that the rule that best describe the translation of the point A(-1,4) is the following:

(x,y) → (x+5,y-3)

Then, the point translated is:

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3 years ago
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5 0
3 years ago
5^(2x-1)+5^(x+1)=250<br> how do you solve?<br> thank you
zepelin [54]

Answer:

<em>x = 2</em>

Step-by-step explanation:

<u>Exponential Equations</u>

Solve:

5^{2x-1}+5^{x+1}=250

Separate each exponential:

5^{2x}5^{-1}+5^{x}5^{1}=250

Operating:

\displaystyle \frac{5^{2x}}{5}+5^{x}5=250

Multiplying by 5:

5^{2x}+25\cdot5^x=1250

Rearranging:

5^{2x}+25\cdot5^x-1250=0

Recall that:

5^{2x}=(5^{x})^2

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Calling

y=5^{x}:

y^2+25y-1250=0

Factoring:

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There are two possible solutions:

y=25

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