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Katarina [22]
1 year ago
6

When we make inferences about the difference of two independent population proportions, what assumptions do we need to make? mar

k all that apply.
Mathematics
1 answer:
Elis [28]1 year ago
4 0

When we make inferences about the difference of two independent population proportions, we assume that it is a random sample, and the number of successes and failures are at least 15 in each group.

Two independent proportions tests involve comparing the proportions of two unrelated datasets.

For these two datasets to be regarded as an independent population, the following must be true or assumed to be true

  • The datasets must represent a random sample
  • Each dataset must contain at least 15 successes and failures

Hence, the above highlights are the assumptions of two independent population proportions.

To learn more about independent populations from the given link

brainly.com/question/23989150

#SPJ4

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Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1
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Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

x'=\frac{dx}{dt}

a)  x’=t–sin(t),  x(0)=1

dx=(t-sint)dt

Apply integral both sides:

\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k

where k is a constant due to integration. With x(0)=1, substitute:

1=0+cos0+k\\\\1=1+k\\k=0

Finally:

x=\frac{t^2}{2} +cos(t)

b) x’+2x=4; x(0)=5

dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\

Completing the integral:

-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}

Solving the operator:

-\frac{1}{2}ln(4-2x)=t+k

Using algebra, it becomes explicit:

x=2+ke^{-2t}

With x(0)=5, substitute:

5=2+ke^{-2(0)}=2+k(1)\\\\k=3

Finally:

x=2+3e^{-2t}

c) x’’+4x=0; x(0)=0; x’(0)=1

Let x=e^{mt} be the solution for the equation, then:

x'=me^{mt}\\x''=m^{2}e^{mt}

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m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i

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x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

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x=\frac{1}{2} sin(2t)

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What is the solution to the equation?
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Answer:

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