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liq [111]
3 years ago
10

Let C be the boundary of the region bounded by x = y4 + 1 and x = 2, oriented in anticlockwise.

Mathematics
1 answer:
murzikaleks [220]3 years ago
4 0
Assuming the first term is \dfrac{e^y}x, we have by Green's theorem,

\displaystyle\int_C\left(\frac{e^y}x\,\mathrm dx+(e^y\ln x+2x)\,\mathrm dy\right)=\iint_R\left(\dfrac\partial{\partial x}\left(e^y\ln x+2x\right)-\dfrac\partial{\partial y}\left(\frac{e^y}x\right)\right)\,\mathrm dA
=\displaystyle\int_{y=-1}^{y=1}\int_{x=y^4+1}^{x=2}\left(\frac{e^y}x+2-\frac{e^y}x\right)\,\mathrm dx\,\mathrm dy
=\displaystyle2\int_{-1}^1\int_{y^4+1}^2\,\mathrm dx\,\mathrm dy
=\displaystyle2\int_{-1}^1(1-y^4)\,\mathrm dy
=\displaystyle4\int_0^1(1-y^4)\,\mathrm dy
=\dfrac{16}5
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Answer:

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=12.8

The sample deviation calculated s=4.324 \approx 4.3

12.8-4.604\frac{4.324}{\sqrt{5}}=3.896    

12.8+ 4.604\frac{4.324}{\sqrt{5}}=21.704    

So on this case the 99% confidence interval would be given by (3.896;21.704)    

Step-by-step explanation:

Data: 10,9,20,13,12

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=12.8

The sample deviation calculated s=4.324

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=5-1=4

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,4)".And we see that t_{\alpha/2}=4.604

Now we have everything in order to replace into formula (1):

12.8-4.604\frac{4.324}{\sqrt{5}}=3.896    

12.8+ 4.604\frac{4.324}{\sqrt{5}}=21.704    

So on this case the 99% confidence interval would be given by (3.896;21.704)    

6 0
3 years ago
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