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:
D. the incenter
Step-by-step explanation:
Hope this helps
15=-4+z/3
+4 +15 = +4 -4 +z/3
cancels the 4 on the right side
19 = z/3
3 x 19 = z/3 x 3
cancels 3 on the right side
I think the answer is z = 57
2^3 is equivalent because if you count the number of twos there are in the parentheses its 3.
The way you find profit is to subtract the revenue and the cost
Profit = Revenue - Cost
The revenue is the amount of money coming in, the cost is the amount of money going out. The goal of course is to have the revenue larger than the cost so that the profit is positive.
So the equation given is
P = 7.5n - (2.25n+15)
and its in the form
P = R - C
where...
R = 7.5n is the revenue equation
C = 2.25n+15 is the cost equation
Focus on the revenue equation
R = 7.5n
which is the same as
R = 7.50*n
This tells us that Sandra pulls in a total of 7.50*n dollars where n is some positive whole number. It represents the number of necklaces sold. For example, if she sold n = 10 necklaces, then
R = 7.50*n
R = 7.50*10
R = 750
meaning that Sandra has made $750 in revenue
As you can see above, the revenue is computed by multiplying the price per necklace ($7.50) by the number of necklaces sold (n) to get R = 7.50*n
So that's why the answer is $7.50
Note: The 2.25 is part of the cost equation. This is known as the variable cost. It is the cost to make one necklace ignoring the fixed cost (eg: rent). The variable cost often doesn't stay the same, but algebra textbooks often simplify this aspect.
