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scoray [572]
3 years ago
5

I don't btgcygljhbihlihnjknkl;n

Mathematics
2 answers:
Digiron [165]3 years ago
7 0
Yeah honestly same it’s ok tho
jekas [21]3 years ago
6 0
Homework do be like that
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Round your answer to the nearest hundredth.<br> ?<br> с<br> А<br> 3<br> 70°<br> B
Paladinen [302]

I'd suggest getting a Texas Instruments TI-84 plus calculator or use des mos .com scientific

4 0
2 years ago
Solve the following inequality:
ad-work [718]

Answer:

The answer is option 3.

Step-by-step explanation:

In order to solve the inequality, you have to get rid of -3 by adding 3 to both sides :

k - 3 \leqslant 9

k - 3 + 3 \leqslant 9 + 3

k \leqslant 12

4 0
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

7 0
3 years ago
Plz help me because I don’t no it<br><br> 2345 ➗ 550
babymother [125]

Answer:

4.26363636364 or this fraction 4 and 6590909090/25000000000

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Do number 1 plz and thank u
Paladinen [302]
It would be A. R= 50 +10m
8 0
3 years ago
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