Answer:
"C" =14
Step-by-step explanation:
14+(-12)=-12+c
-12+c=14+(-12)
-12+c=14-12
-12+c+12=2+12
c=14
7.) Area of a square pyramid is given by A = s^2 + 2sl; where s is the side length of the base and l is the slant height.
Area of given pyramid = 4^2 + (2 x 4 x 7) = 16 + 56 = 72 ft^2
8.) Area of the given pyramid is the area of the hexagonal base plus the area of the six slant triangles.
Area of hexagonal base = (3sqrt(3))/2 x 10^2 = 150 x 1.732 = 259.8
Area of the 6 traingles = 6(1/2 x 10 x 13) = 6 x 65 = 390
Total surface area of the pyramid = 259.8 + 390 = 649.8 ≈ 650 m^2
9.) Using pythagoras rule, the slant height is sqrt(6^2 + 3.5^2) = sqrt(36 + 12.25) = sqrt(48.25) = 6.9 mm
10.) The surface area of a cone is given by πr^2 + πrl
Area = π x 6^2 + π x 6 x 20 = 36π + 120π = 490.1 cm^2
11.) l = sqrt(r^2 + h^2) = sqrt(11^2 + 16^2) = sqrt(121 + 256) = sqrt(377) = 19m
Answer:
I'm assuming you have 4 options, and if so, the answer is A.
Step-by-step explanation:
Rounding 36.6 off to 37 will make it easier to estimate the quotient when dividing by 5. For example, I can easily come up with a solution in my head when dividing 37 to 5 but it's harder for me to come up with a solution to 36.6.
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)
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:
Answer:
you need a d/protector to do this
Step-by-step explanation:
