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soldier1979 [14.2K]
3 years ago
8

What is the equation of the epicycloid formed by a circle with radius 6 rolling counterclockwise around a circle with radius 12

Mathematics
1 answer:
Ilya [14]3 years ago
3 0

Answer:

Step-by-step explanation:

Radius=24$;*3)5435)

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I don’t know how to do it, please teach me !!!
cricket20 [7]

As described in z-distribution the answers are given below:

a) The 95% confidence interval estimate for the population mean spending by a customer on visiting salon per month is given as follows: (747, 853).

b) The sampling error at 95% confidence level is of: $35.78.

What is a z-distribution ?

The normal distribution with a mean of 0 and a standard deviation of 1 is referred to as the standard normal distribution (also known as the Z distribution) (the green curves in the plots to the right). It is frequently referred to as the bell curve since the probability density graph resembles a bell.

solution:

The bounds of the confidence interval are given as follows:

In which:

is the sample mean.

z is the critical value.

n is the sample size. is the standard deviation for the population.

The parameters for this problem are given as follows:

Hence the lower bound of the interval is of:

800 - 200 x 1.96/square root of 55  = 747.

The upper bound of the interval is of:

800 + 200 x 1.96/square root of 55  = 853.

The sampling error for a sample size of 120 is calculated as follows:

200 x 1.96/square root of 120 = $35.78.

To learn more about the z-distribution from the given link

brainly.com/question/4079902

#SPJ1

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1 year ago
Hypothetically, if you possess a numerical value of two, and gain an additional numerical value of two, what would be the comple
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267.3
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3 years ago
I’m not sure on what to do for these problems…can someone help me please
Ray Of Light [21]

Answer:

You need to find k, the constant of proportionality, by dividing y with its corresponding x.

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lim x rightarrow 0 1 - cos ( x2 ) / 1 - cosx The limit has to be evaluated without using l'Hospital'sRule.
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Answer with Step-by-step explanation:

Given

f(x)=\frac{1-cos(2x)}{1-cos(x)}\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-(cos^2{x}-sin^2{x})}{1-cos(x)})\\\\(\because cos(2x)=cos^2x-sin^2x)\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-cos^2x}{1-cos(x)}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}(\frac{(1-cosx)(1+cosx)}{1-cosx}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}((1+cosx)+\frac{sin^2x}{1-cosx})\\\\\therefore \lim_{x \rightarrow 0}f(x)=1

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