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netineya [11]
1 year ago
6

Find the indefinite integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) dx x(l

n x2)5
Mathematics
1 answer:
Pepsi [2]1 year ago
7 0

I'm assuming the integral is

\displaystyle \int \frac{dx}{x (\ln(x^2))^5}

We have

\ln(x^2) = 2 \ln|x| \implies (\ln(x^2))^5 = 32 (\ln|x|)^5

Then substituting y=\ln|x| and dy=\frac{dx}x, the integral transforms and reduces to

\displaystyle \int \frac{dx}{x(\ln(x^2))^5} = \frac1{32} \int \frac{dy}{y^5} \\\\ ~~~~~~~~ = \frac1{32} \left(-\frac1{4y^4}\right) + C \\\\ ~~~~~~~~ = -\frac1{128(\ln|x|)^4} + C

which we can rewrite as

128 (\ln|x|)^4 = 8\cdot2^4(\ln|x|)^4 = 8 (2\ln|x|)^4 = 8 (\ln(x^2))^4

and so

\displaystyle \int \frac{dx}{x (\ln(x^2))^5} = \boxed{-\frac1{8(\ln(x^2))^4} + C}

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Greatest common factor for 32-4x
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At a grocery store, blueberries come packaged in 8-ounce containers for $2.80. At a farmer's market, blueberries cost $4.20 for
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The answer is letter C.

In the grocery, the prices per product is represented by: 
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The farmer's market blueberries is cheaper than the grocery's by $0.80 per pound.
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These data can be approximated quite well by a N(3.4, 3.1) model. Economists become alarmed when productivity decreases. Accordi
CaHeK987 [17]

Answer:

First part

P(X< 3.4-0.6*3.1) = P(X

And for this case we can use the z score formula given by:

z = \frac{x- \mu}{\sigma}

And using this formula we got:

P(X

And we can use the normal standard table or excel and we got:

P(Z

Second part

For the other part of the question we want to find the following probability:

P(-1.715

And using the score we got:

P(-1.715

And we can find this probability with this difference:

P(-1.65< Z< 1.210)=P(Z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the data of a population, and for this case we know the distribution for X is given by:

X \sim N(3.4,3.1)  

Where \mu=3.4 and \sigma=3.1

First part

And for this case we want this probability:

P(X< 3.4-0.6*3.1) = P(X

And for this case we can use the z score formula given by:

z = \frac{x- \mu}{\sigma}

And using this formula we got:

P(X

And we can use the normal standard table or excel and we got:

P(Z

Second part

For the other part of the question we want to find the following probability:

P(-1.715

And using the score we got:

P(-1.715

And we can find this probability with this difference:

P(-1.65< Z< 1.210)=P(Z

8 0
3 years ago
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