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meriva
3 years ago
10

Need some help Appreciate any help Take a look at the screenshot

Mathematics
2 answers:
malfutka [58]3 years ago
8 0

Answer:

y = -1/2x+2

Step-by-step explanation:

Two points on the line are (0,2) and (4,0)

We can find the slope

m = (y2-y1)/(x2-x1)

   = (0-2)/(4 -0)

   = -2/4

    - 1/2

We know the y intercept is 2 since it crosses the y axis at 2

The slope intercept form is

y = mx+b where m is the slope and b is  the y intercept

y = -1/2x+2

katrin2010 [14]3 years ago
7 0

Answer:

See below.

Step-by-step explanation:

First, from the graph, we can already pick out two points: (0,2) and (4,0). Note that (0,2) is the y-intercept.

Therefore, we can use slope-intercept form or y=mx+b where m is the slope and b is the y-intercept.

We already know the b, so:

y=mx+2

Now, we can plug in the ordered pair (4,0) to help us determine the slope (or you can just use the slope formula for the two ordered pairs):

0=m(4)+2

-2=4m

m=-1/2

Thus, the entire equation is:

y=-\frac{1}{2} x+2

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Step-by-step explanation:

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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:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

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

And we can write this as an infinite series like this:

\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}

And then we have this:

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

And then we can find the E[X^2]

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

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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For this case we have the following system of equations:

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Thus, we have the equivalent system:

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We add the equations:

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Answer:

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