A simple way to reduce noise is to perform what is called a moving average filter on a signal. Mathematically, the moving average filter is a discrete-time random process described by Y _n = 1/L sigma^n_k = n - L + 1 X_k
where X_k are the original input random process needing noise reduction. Assume that the process X_k is IID and has mean 2 and variance 3.
a. Find the mean and variance of Y_n for L = 12.
b. Based on the Central Limit Theorem, write an expression for the PDF of Y_n.
c. How large should L be to have the variance Y_n be less than 0.1?
d. For the value of L computed in part (c), what is Pr [|Y_n - E{Y_n}|] > 0.1?

isi

step-by-step explanation:

use that smart brain

50- 250 is correct i did that one too

step-by-step explanation:

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.