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Setler [38]
2 years ago
8

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Evgen [1.6K]2 years ago
8 0

Answer:

3.3 and 5

Step-by-step explanation:

Suppplement angle is totaling both of angle to 180 degree

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Please asap I need this for a test
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Answer:

3

Step-by-step explanation:

(f O g) this basically means that the input x first goes trough the function g and then f. Like f(g(x)).

So when x went trough g, you got the output g(x) and then this went trough f and you got f(g(x)) = -8 = 'f(x)'.

With this in mind you can retrace your steps by first looking at what input can get -8 as an output, for f this is -4. this means g(x) = -4

Then you look at what input (this is the x you're looking for) gets you the ouput -4. Looking at the second image you'll picture see that it's the input 3.

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2 years ago
If michelle walked 5 meters, how many inches did she walk
DerKrebs [107]
1 meter = 39.370 inches.....so 5 meters = (5)(39.370) = 196.85 inches
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Does this graph show a function?explain how you know.
DiKsa [7]
The answer is B! It passes the vertical line test
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3 years ago
Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
LenaWriter [7]

Answer:

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

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

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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3 years ago
9. Select a set of corresponding angles.
Anettt [7]

Answer:

A. Angles M and D

3 0
3 years ago
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