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3 years ago

I do not understand at all and my answer is probably completely wrong please help

Mathematics

2 answers:

Stolb23 [73]3 years ago

6 0

Your answer is correct but for your top answer 12/12 equals 1 so just put one.

tatuchka [14]3 years ago

4 0

25 minutes!! you were very close, you just needed to translate the fraction to minutes!

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HELP ASAP WILL GIVE BRAINLIEST!!!!!

djyliett [7]

0.32. Because you need to find the area, as size 8*8/2 then divide by 100 to get the decimal.I hope this helps.

5 0

3 years ago

X 50 50 X = degrees Please help with this problem I am going crazy

sveta [45]

Answer:

x=80°

Step-by-step explanation:

Triangles always add up to 180°.

As you have 2 angles, you would have to add them together and then take it away from 180° to get your answer.

50+50=100°

180-100=80°

Hope this helps :)

6 0

2 years ago

Read 2 more answers

1. Write the following radicals in simplest form a) -3√288 b) 5√320

jeka57 [31]

A) -3√288=-3√(144*2)=-3√144√2=-3*12√2=-36√2

b) 5√320=5√(64*5)=5√64√5=5*8√5=40√5

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3 years ago

What is the slope and y-intercept of a vertical line

Dmitry_Shevchenko [17]

Is there any numbers or a picture

7 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