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siniylev [52]
3 years ago
5

John says the transformation rule (x, y) > (x + 4, y + 7) can be used to describe the slide of the pre-image (4, 5) to the im

age (0, –2). What was his error?
Mathematics
1 answer:
erik [133]3 years ago
5 0

If he used the rule (x + 4, y + 7) then pre-image (4, 5) would result in image (4 + 4, 5 + 7) = (8, 12).  

He should have used the rule (x - 4, y - 7)


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Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
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Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

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Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

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\phi(t) = e^{\frac{t^2}{2}

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\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

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E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

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And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

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