Someone help me please

Hey ! Can anyone help me solve these problems please. Thank you !

devlian [24]

The answer to solve these problems is that the full circle is 360 degrees, the way to find the answer is to have to equal sides that equal 360 degrees. That means one half of 360 degrees is 180 degrees. Hope this helps.

6 0

2 years ago

X -1 < -4<br><br> What is ittttttttt

Kamila [148]

Answer:

x<-3

Step-by-step explanation:

x-1+1<-4+1

x<-3

8 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

How do i draw two number lines that show 0.20 and 1/5 are equivalent?

leonid [27]

The frist line may be one with the numbers 0, 1 and 2 and 10 divisions (marks at equal distance) between each integer. Each division will be equal to 0.1 units and then you can mark the second division from the zero point to the right as the 0.20 mark.

The other line must have the same integers, 0 , 1 and 2 placed in identical form as the first line. Then
   - draw an inclined straight line since the point zero,
   - mark 5 points in the inclined lined equally spaced over the line.
   - draw a sttraight line from the 5th point to the point with the mar 1 over the base number line.
   - draw a parallel line to the previous one passing trhough the second point of the inclined line and mark the point where this parallel touchs the base number line. This point shall be at the same distance from zero than the 0.2 mark was in the first number line, meaning that 0.2 and 1/5 are equivalent.

6 0

3 years ago

Your business sells cupcakes in boxes of 10. The demand equation is

melomori [17]

Step by step equation

5 0

3 years ago

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