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:
x= -1.5, y= -1
Step-by-step explanation:
6x= -5+4y
x= (-5+4y) / 6
4((-5+4y) / 6) + 5y = -11
4(-5+4y) + 5*6y = -11*6
-20+16y+30y= -66
16y + 30y = -66+20 = -46
y(16 + 30) = -46
y= -46 / (16 + 30)
y= -46 / 46
y= -1
x= (-5+4(-1)) / 6
x= (-5-4) / 6
x= -9 /6
x= -1.5
Answer: 9 in
Step-by-step explanation:
Answer:
Step-by-step explanation:
What is the question?
g(x) = -0.5x² + 4x - 2 is a down-opening parabola. Do you need to know how to put it in vertex form?
vertex (4,6)
focus (4,5.5)
