Let X Y Be Compact Metric Spaces And Consider The Metric Space X X Y Equipped Wi

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Please help me prove the following about sequences of continuous functions

Let X, Y be compact metric spaces and consider the metric space X x Y equippedwith the metric d((x1, y1), (x2, y2)) = dx(x1, 72) + dy (y1, y2). Prove that for everyf EC(X xY) and every e > 0, there exist functions gi, . . . On E C(X) and hi, . .. hn EC(Y) such thatnIf (x, y) -gi (x )hi(y) <e for all x EX, yEY.j=1

 
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Let X Be A Continuous Random Variable With Cdf F X Taking Values In An Interval

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4. Let X be a continuous random variable, with CDF F(x), taking values in an interval[0, b]; that is, F(0) = 0 and F(b) = 1. Then there is an alternative formula forexpected value:E(X) =Z b0(1 − F(x)) dx. (1)(a) Assume b is a finite number. Prove (1) using integration-by-parts. [Hint:Recall that the PDF is f(x) = ddxF(x).](b)Check that the formula (1) holds when X Unif(0, b).(c) Formula (1) also works for b = 1. Check this when X is an Exponential RVwith PDF f(x) = e−x for x 0.

 
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Let X X U V U V2 And Y Y U V U2 V4

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5. Let x = x(u, v) = u + v2, and y = y(u, v) = u2 + v4.(a). Find out where the Jacobian of the transformation equals 0, and find the chambers in the u-v plane where the Jacobian is positive and the chambers where it is negative.(b). Find the curves in the x-y plane that correspond to the rays in the u-v plane where the Jacobian equals 0. (One of the curves is only half a curve!)(c). As you go counterclockwise about the origin in the u-v plane, describe the correspondingmotion in the x-y plane, using your answer to part (a).

 
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Let X Bar Be The Mean Of A Random Sample From The Exponential Distribution

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Let X bar be the mean of a random sample from the exponential distribution.

a) show that xbar is unbiasedpoint estimator of  θ

b) using the mgf technique,determine the distribution of xbar

c) use (b) to show that Y= 2nx bar/ θ has x^2 (chi) distibution with 2n degrees of freedom.

d) based on part (c),find a 95 % confidence interval for  θ if n=10

 
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Let X U K 1 K I For Any N E N Prove That I X 1 Z By Lling Out The Gaps In The 15

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Please help with the attached question. Don’t have to rewrite the question, it’s just fill the blanks

Let X,, = U {k —;1;,k+ i) for any n E N. Prove that I] X,1 = Z by filling out the gaps in the#152 116"argument below. Proof. For any n E N and 1′ E R we have _< :1: <_. In particular, if :1: E Z, then :r E __ forany n E N as _. Therefore, E C X,1 for any n E N. As a result, _. We want to show that n X“ C Z. From the equivalence of the “if-then” statement and itsnEN contrapositive, it is enough to show that _. Notice that n Xn C IR as for all k E Z and allHEN n E N we have _. Therefore, we want to show that if 1: E_, then _. Assume r E_. Then,choose k E E such that k < r < k + 1. In particular, a.“ — k_ and (k +1)— :t_, so we canchoose or E N such that _ and _ because _. As a result, a >_ and a: <_. Therefore, 1: 5-3 _. So, :2: ¢_. Since _ and _, we have _. III

 
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Let X And Y Be Two Correlated Random Variables If The Correlation Coefficient Be

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Let X and Y be two correlated random variables. If the correlation coefficient between X

and Y is1 ,i.e., ρ(X,Y)=1, show that

Question:Let X and Y be two correlated random variables. If the correlation coefficient between Xand Y is 1.Answer:p ( X ,Y ) = COV ( X , Y )√V ( X )∗V ( Y ) Since , p ( X ,Y )=1, V(X+Y) =…

 
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Let X T Be A Signal That Has A Rational Laplace Transform With Exactly Two Poles

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Let x(t) be a signal that has a rational Laplace transform with exactly two poles, located at s = -1 and s = -3. If g(t) = e2t x(t) and G (jw) [ the Fourier transform of g(t)] converges, determine whether x(t) is left sided, right sided, or two sided?

 
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Let X And Y Be Random Variables With E X Mx 40 Sd X Ox 10 E Y Hy 71 Sd Y Oy 2 Co

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Need help for a and b. Here is the question. Thank you!

Let X and Y be random variables withE ( X ) = MX = 40 , SD ( X ) = OX = 10,E ( Y ) = HY = 71, SD ( Y ) = OY = 2, Cor ( X , Y ) = P = – 0. 9 .a . Find E ( 2 X – 7Y ) and SD ( 2 X – 7Y ) . Give your answers to 2 decimal places .E ( 2 X – 7Y ) =SD ( 2 X – 7Y ) =Submit AnswerIncorrect . Tries 1 / 5 Previous Triesb . Let X and Y be random variables withSD ( X ) = OX = 2, SD ( Y ) = OY = 9, SD ( 2 X + 3 Y ) = 28.Find Cor ( X , Y ) = P. Give your answer to 4 decimal places .P =Submit Answer`Tries 0 / 5

 
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Let X Represent The Number Of Mountain Climbers Killed Each Year The Long Term V 1

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xxσ26s2116.2

(a) What is the level of significance? 

State the null and alternate hypotheses.

Ho: σ2 < 136.2; H1: σ2 = 136.2

Ho: σ2 = 136.2; H1: σ2 ≠ 136.2    

Ho: σ2 = 136.2; H1: σ2 < 136.2

Ho: σ2 = 136.2; H1: σ2 > 136.2

(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.) 

What are the degrees of freedom? 

What assumptions are you making about the original distribution?

We assume a normal population distribution.

We assume a exponential population distribution.    

We assume a binomial population distribution.

We assume a uniform population distribution.

(c) Find or estimate the P-value of the sample test statistic. 

P-value > 0.100

0.050 < P-value < 0.100    

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value > α, we fail to reject the null hypothesis.

Since the P-value > α, we reject the null hypothesis.    

Since the P-value ≤ α, we reject the null hypothesis.

Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 1% level of significance, there is insufficient evidence to conclude that the variance for number of mountain climber deaths is less than 136.2

At the 1% level of significance, there is sufficient evidence to conclude that the variance for number of mountain climber deaths is less than 136.2    

(f) Find the requested confidence interval for the population variance. (Round your answers to two decimal places.)lower limit 

upper limit     

Interpret the results in the context of the application.

We are 90% confident that σ2 lies outside this interval.

We are 90% confident that σ2 lies below this interval.    

We are 90% confident that σ2 lies above this interval.

We are 90% confident that σ2 lies within this interval.

 
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Let X Represent The Dollar Amount Spent On Supermarket Impulse Buying In A 10 Mi

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xx $27$8

(a) Consider a random sample of n = 40 customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say about the probability distribution of x, the average amount spent by these customers due to impulse buying? What are the mean and standard deviation of the x distribution?

The sampling distribution of x is not normal.

The sampling distribution of x is approximately normal with mean μx = 27 and standard error σx = $8.    

The sampling distribution of x is approximately normal with mean μx = 27 and standard error σx = $0.20.

The sampling distribution of x is approximately normal with mean μx = 27 and standard error σx = $1.26.

Is it necessary to make any assumption about the x distribution? Explain your answer.

It is not necessary to make any assumption about the x distribution because μ is large.

It is necessary to assume that x has an approximately normal distribution.    

It is necessary to assume that x has a large distribution.

It is not necessary to make any assumption about the x distribution because n is large.

(b) What is the probability that x is between $24 and $30? (Round your answer to four decimal places.)

(c) Let us assume that x has a distribution that is approximately normal. What is the probability that x is between $24 and $30? (Round your answer to four decimal places.)

(d) In part (b), we used x, the average amount spent, computed for 40 customers. In part (c), we used x, the amount spent by only one customer. The answers to parts (b) and (c) are very different. Why would this happen?

The standard deviation is smaller for the x distribution than it is for the x distribution.

The sample size is smaller for the x distribution than it is for the x distribution.    

The standard deviation is larger for the x distribution than it is for the x distribution.

The mean is larger for the x distribution than it is for the x distribution.

The x distribution is approximately normal while the x distribution is not normal.

In this example, x is a much more predictable or reliable statistic than x. Consider that almost all marketing strategies and sales pitches are designed for the average customer and not the individual customer. How does the central limit theorem tell us that the average customer is much more predictable than the individual customer?

The central limit theorem tells us that small sample sizes have small standard deviations on average. Thus, the average customer is more predictable than the individual customer.

The central limit theorem tells us that the standard deviation of the sample mean is much smaller than the population standard deviation. Thus, the average customer is more predictable than the individual customer.  

 
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