72
107-9=98
98-78=90
90-7=83
83-6=7
77-5=72
72-4=68
68-3=65
Answer:
Cov(X, Y) =0.029.
Step-by-step explanation:
Given that :
The noise in a particular voltage signal has a constant mean of 0.9 V. that is μ = 0.9V ............(1)
Also, the two noise instances sampled τ seconds apart have a bivariate normal distribution with covariance.
0.04e–jτj/10 ............(2)
Having X and Y denoting the noise at times 3 s and 8 s, respectively, the difference of time = 8-3 = 5seconds.
That is, they are 5 seconds apart,
τ = 5 seconds..............(3)
Thus,
Cov(X, Y), for τ = 5seconds = 0.04e-5/10
= 0.04e-0.5 = 0.04/√e
= 0.04/1.6487
= 0.0292
Thus, Cov(X, Y) =0.029.
Answer:
yes
Step-by-step explanation:
Answer:
see explanation
Step-by-step explanation:
let y = f(x), then
y = 11x - 1
Interchange x and y and solve for y
x = 11y - 1 ( add 1 to both sides )
x + 1 = 11y ( divide both sides by 11 )
y = 
, hence
(x) =
176 divided by 8 = 22
So 176 can be evenly divided by 8
