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koban [17]
3 years ago
13

A sampling frame a. is a list of population elements from which the sample will be drawn. b. is the list of population elements

actually included in the sample. c. usually provides biased statistics. d. is a form of probability sampling.
Mathematics
1 answer:
olga2289 [7]3 years ago
6 0

Answer:

<h2> A. is a list of population elements from which the sample will be drawn</h2>

Step-by-step explanation:

Sampling frame is the list of people or items from which a sample is taken.  

Sampling is used because it is almost impossible to study entire population. Sampling allows us to find information about a population on the basis of results from a subset of the population, without studying every individual. It is easier and cost effective process.

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You and three friends go to the town carnival. You have a coupon for $20 off that will save your group money! If the total bill
Lunna [17]

Answer:

30 that singular tickets for each friend and if your asking for total it would be 120. :)

Step-by-step explanation:

They paid 100 but they had a 20$ off coupon, this means the orignial total would be $120 so 120 divided by 4 (there's four people) = $30 each :)

hope this helped

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Find the point-slope equation for
mars1129 [50]

Answer:

y-24=m(x--4)

Step-by-step explanation:

point slope form:

y-y₁=m(x-x₁)

1) find the slope or "m" first:

slope formula: (y2-y1)/(x2-x1)

(-53-24)/7+4

-77/11=-7

slope=-7

so

y-y₁=-7(x-x₁)

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y-24=m(x--4)

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8 0
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What is the ratio 44/12 in simplest form
sladkih [1.3K]

Answer:

44/12 = 11/3 =  3   2/3

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
What is the antiderivative of 3x/((x-1)^2)
Maslowich

Answer:

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Step-by-step explanation:

Given

\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx

\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx

=3\cdot \int \frac{x}{\left(x-1\right)^2}dx

\mathrm{Apply\:u-substitution:}\:u=x-1

=3\cdot \int \frac{u+1}{u^2}du

\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}

=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx

=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)

as

\int \frac{1}{u}du=\ln \left|u\right|     ∵ \mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)

\int \frac{1}{u^2}du=-\frac{1}{u}        ∵     \mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1

so

=3\left(\ln \left|u\right|-\frac{1}{u}\right)

\mathrm{Substitute\:back}\:u=x-1

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)

\mathrm{Add\:a\:constant\:to\:the\:solution}

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Therefore,

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

4 0
3 years ago
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