Answer:
100 degrees
Step-by-step explanation:
The angle x is opposite the angle that is 100 degrees.
Since they are opposite each other, they are vertical angles.
This means that they are congruent, or equal to each other.
Therefore,
x=100
x is 100 degrees
The sound wave with a <u>frequency of 20</u> waves/sec is 800 longer than the wavelength of a sound wave with a <u>frequency of 16,000</u> waves/sec
<h3>Calculating wavelength </h3>
From the question, we are to determine how many times longer is the first sound wave compared to the second sound water
Using the formula,
v = fλ
∴ λ = v/f
Where v is the velocity
f is the frequency
and λ is the wavelength
For the first wave
f = 20 waves/sec
Then,
λ₁ = v/20
For the second wave
f = 16,000 waves/sec
λ₂ = v/16000
Then,
The factor by which the first sound wave is longer than the second sound wave is
λ₁/ λ₂ = (v/20) ÷( v/16000)
= (v/20) × 16000/v)
= 16000/20
= 800
Hence, the sound wave with a <u>frequency of 20</u> waves/sec is 800 longer than the wavelength of a sound wave with a <u>frequency of 16,000</u> waves/sec
Learn more on Calculating wavelength here: brainly.com/question/16396485
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Answer:
![E[X^2]= \frac{2!}{2^1 1!}= 1](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%20%5Cfrac%7B2%21%7D%7B2%5E1%201%21%7D%3D%201)

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}]](https://tex.z-dn.net/?f=%20%5Cphi%28t%29%20%3D%20E%5Be%5E%7BtX%7D%5D)
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:

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

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

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 +....]](https://tex.z-dn.net/?f=E%5Be%5E%7BtX%7D%5D%3D%20E%5B1%2B%20tX%20%2B%5Cfrac%7B1%7D%7B2%7D%20%28tX%29%5E2%20%2B....%2B%5Cfrac%7B1%7D%7Bn%21%7D%28tX%29%5En%20%2B....%5D)
![E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...](https://tex.z-dn.net/?f=E%5Be%5E%7BtX%7D%5D%3D%201%2B%20E%5BX%5Dt%20%2B%5Cfrac%7B1%7D%7B2%7DE%5BX%5E2%5Dt%5E2%20%2B....%2B%5Cfrac%7B1%7D%7Bn1%7DE%5BX%5En%5D%20t%5En%2B...)
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}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%282k%29%21%7D%20E%5BX%5E%7B2k%7D%5D%20t%5E%7B2k%7D%3D%5Cfrac%7B1%7D%7Bk%21%7D%20%28%5Cfrac%7Bt%5E2%7D%7B2%7D%29%5Ek%20%3D%5Cfrac%7B1%7D%7B2%5Ek%20k%21%7D%20t%5E%7B2k%7D)
And then we have this:
![E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...](https://tex.z-dn.net/?f=E%5BX%5E%7B2k%7D%5D%3D%5Cfrac%7B%282k%29%21%7D%7B2%5Ek%20k%21%7D%2C%20k%3D0%2C1%2C2%2C...)
And then we can find the ![E[X^2]](https://tex.z-dn.net/?f=E%5BX%5E2%5D)
![E[X^2]= \frac{2!}{2^1 1!}= 1](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%20%5Cfrac%7B2%21%7D%7B2%5E1%201%21%7D%3D%201)
And we can find the variance like this :
![Var(X^2) = E[X^4]-[E(X^2)]^2](https://tex.z-dn.net/?f=Var%28X%5E2%29%20%3D%20E%5BX%5E4%5D-%5BE%28X%5E2%29%5D%5E2)
And first we find:
![E[X^4]= \frac{4!}{2^2 2!}= 3](https://tex.z-dn.net/?f=E%5BX%5E4%5D%3D%20%5Cfrac%7B4%21%7D%7B2%5E2%202%21%7D%3D%203)
And then the variance is given by:

9514 1404 393
Answer:
B, F
Step-by-step explanation:
The volume is the product of the dimensions:
V = LWH
V = (b^2)(3c^2)(2c^1) = (3·2)(b^2)(c^(2+1))
V = 6b^2·c^3 . . . . . matches B
__
For b=3 and c=2, the volume is ...
V = 6(3^2)(2^3) = 6(9)(8)
V = 432 . . . . cubic inches . . . . matches F
The given expression is

We have to write it in a different way and for that, we have to check the factors of 125 and x^4 .
And factors of 125 are 25 and 5.

Radical 25 equals to 5 and radical
is
.
So we will get

And that's the another way to write the given expression .