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ryzh [129]
3 years ago
6

Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th

ink of the corresponding vector solutions y1 and y2 and use the Wronskian to show that the solutions are linearly independent Wronskian = det [ ] = These solutions are linearly independent because the Wronskian is [ ] Choose for all x.
Mathematics
1 answer:
xenn [34]3 years ago
3 0

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is \bold{W=-3e^{7x}}

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|

Thus replacing the functions of the exercise we get:

W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|

Working with the determinant we get

W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

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Answer:

a) Null Hypothesis: \mu =900

Alternative hypothesis: \mu \neq 900

b) The 95% confidence interval would be given by (910.05;959.95)    

c) Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d) z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750

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p_v =2*P(Z>2.750)=0.0059

Step-by-step explanation:

a. State the hypotheses.

On this case we want to check the following system of hypothesis:

Null Hypothesis: \mu =900

Alternative hypothesis: \mu \neq 900

b. What is the 95% confidence interval estimate of the population mean examination  score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

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=935 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=180 represent the population standard deviation

n=200 represent the sample size  

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

\bar X \pm z_{\alpha/2}\frac{\sigma}{\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=3278.222

The sample deviation calculated s=97.054

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

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

935-1.96\frac{180}{\sqrt{200}}=910.05    

935+1.96\frac{180}{\sqrt{200}}=959.95    

So on this case the 95% confidence interval would be given by (910.05;959.95)    

c. Use the confidence interval to conduct a hypothesis test. Using α= .05, what is your  conclusion?

Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d. What is the p-value?

The statistic is given by:

z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

If we replace we got:

z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750

Since is a bilateral test the p value is given by:

p_v =2*P(Z>2.750)=0.0059

So then since the p value is less than the significance we can reject the null hypothesis at 5% of significance.

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