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3 years ago

What is a cluster sample?

1 answer:

4 0

This is a sampling technique used when natural groupings are evident in a statistical population. It is often used in marketing research.

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Find the indefinite integral. (Note: Solve by the simplest method—not all require integration by parts. Use C for the constant o

umka2103 [35]

Answer:

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2 ln⁡(1+x^2 )+C

Step-by-step explanation:

∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗

Let 1st=arctan⁡(x)

And 2nd=1

∫▒〖arctan⁡(x).1 dx=arctan⁡(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗

As we know that

derivative of arctan(x)=1/(1+x^2 )

∫▒〖1 dx〗=x

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1

Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now

Let 1+x^2=u

du=2xdx

Multiply and divide ∫▒〖(1/(1+x^2 ))dx.x〗 by 2 we get

1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)

1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)

1/2 ∫▒(2xdx/u) =1/2 ln⁡(u)+C

1/2 ∫▒(2xdx/u) =1/2 ln⁡(1+x^2 )+C

Putting values in Eq1 we get

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2 ln⁡(1+x^2 )+C (required soultion)

3 0

3 years ago

Read 2 more answers

What is the value of the following expression? <br> -9/4➕-8/5-2 1/2

SOVA2 [1]

The answer to the expression is -97/20 or -4.85

3 0

3 years ago

svetoff [14.1K]

Answer:

1/2 ornament in an hour

Step-by-step explanation:

Given

$Orna\ ments = 2$

$Time = 4$

Required

Determine the ornaments per hour

This question implies that we calculate the unit rate.

$Unit\ Rate = \frac{Orna\ ments}{Time}$

Substitute values for Time and number of ornaments

$Unit\ Rate = \frac{2}{4}$

$Unit\ Rate = \frac{1}{2}$

<em>This implies that Constance can decorate 1/2 ornaments in an hour</em>

3 0

2 years ago

Simplify the expression: Show your work. Write the answer in standard form (15 +6d - d²) + (-5d² - 18 + 6d) PLEASE HELP

mezya [45]

Answer:

-6d^2 + 12d - 3

Step-by-step explanation:

use PEMDAS

parentheses

exponents

multiplication

division

addition

subtraction

use that order to solve problems like that.

6 0

3 years ago

In March 2007, Business Week reported that at the top 50 business schools, students studied an average of 14.6 hours. You wonder

Rasek [7]

Answer:

The hypotheses used in this situation

$H_0:\mu = 14.6$

$H_a:\mu \neq 14.6$

Step-by-step explanation:

We are given that Business Week reported that at the top 50 business schools, students studied an average of 14.6 hours.

Mean = $\mu = 14.6$

Claim : The amount UMSL students study is different from this 14.6 hour benchmark.

The hypotheses used in this situation

$H_0:\mu = 14.6$

$H_a:\mu \neq 14.6$

5 0

3 years ago