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3 years ago

Solve the differential equation dy/dx=2*y/x,x>0 simplify?

1 answer:

4 0

Separate your variables:

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2y}x\implies \dfrac{\mathrm dy}y=2\dfrac{\mathrm dx}x$

Integrating both sides gives

$\displaystyle\int\frac{\mathrm dy}y=2\int\frac{\mathrm dx}x\implies \ln|y|=2\ln|x|+C\implies y=e^{2\ln|x|+C}=Cx^2$

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The product of a negative 11 cubed and negative 11 squared

bekas [8.4K]

Answer:

121 and 1331

Step-by-step explanation:

Because 11x11=121 and 11x11x11=1331

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3 years ago

Identify the transformation that maps the figure onto itself.

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The answer for this letter is c

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4 years ago

Read 2 more answers

You pay 26.50 for an item including 6% sales tax, what was the original price. Show all work please.

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26.5×.006=value, so you take value ans subtract from 26.5.

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3 years ago

A very large batch of parts (assume a normal distrbution0 from a manufacturer has a mean weight = 43g and a standard deviation =

AVprozaik [17]

Answer:

5.44% probability that exactly 8 of the 16 parts you selected will have weights exceeding 45g

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Binomial probability distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Percentage of parts with weights exceeding 45g?

1 subtracted by the pvalue of Z when X = 45. So

We have $\mu = 43, \sigma = 4$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{45 - 43}{4}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915

1 - 0.6915 = 0.3075

If you slect 16 parts at random form that batch, what is the probability that exactly 8 of the 16 parts you selected will have weights exceeding 45g?

This is P(X = 8) when n = 16, p = 0.3075. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{16,8}.(0.3075)^{8}.(0.6915)^{8} = 0.0544$

5.44% probability that exactly 8 of the 16 parts you selected will have weights exceeding 45g

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3 years ago

Expand to write an equivalent expression: -14(-8x+12y)

makkiz [27]

Answer:

an equivalent expression: -14(-8x+12y)=112x-168y

factor of an equivalent expression: 36a−16=4(9a-4)

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3 years ago