Answer:
a) Var[z] = 1600
D[z] = 40
b) Var[z] = 2304
D[z] = 48
c) Var[z] = 80
D[z] = 8.94
d) Var[z] = 80
D[z] = 8.94
e) Var[z] = 320
D[z] = 17.88
Step-by-step explanation:
In general
V([x+y] = V[x] + V[y] +2Cov[xy]
how in this problem Cov[XY] = 0, then
V[x+y] = V[x] + V[y]
Also we must use this properti of the variance
V[ax+b] = 
V[x]
and remember that
standard desviation = ![\sqrt{Var[x]}](https://tex.z-dn.net/?f=%5Csqrt%7BVar%5Bx%5D%7D)
a) z = 35-10x
Var[z] = 
Var[x] = 100*16 = 1600
D[z] = 
= 40
b) z = 12x -5
Var[z] = 
Var[x] = 144*16 = 2304
D[z] = 
= 48
c) z = x + y
Var[z] = Var[x+y] = Var[x] + Var[y] = 16 + 64 = 80
D[z] = 
= 8.94
d) z = x - y
Var[z] = Var[x-y] = Var[x] + Var[y] = 16 + 64 = 80
D[z] = = 8.94
e) z = -2x + 2y
Var[z] = 4Var[x] + 4Var[y] = 4*16 + 4*64 = 320
D[z] = 
= 17.88
Answer:
The pairs are (13,15) and (-15,-13).
Step-by-step explanation:
If n is an odd integer, the very next odd integer will be n+2.
n+1 is even (so we aren't using this number)
The sum of the squares of (n) and (n+2) is 394.
This means
(n)^2+(n+2)^2=394
n^2+(n+2)(n+2)=394
n^2+n^2+4n+4=394 since (a+b)(a+b)=a^2+2ab+b^2
Combine like terms:
2n^2+4n+4=394
Subtract 394 on both sides:
2n^2+4n-390=0
Divide both sides by 2:
n^2+2n-195=0
Now we need to find two numbers that multiply to be -195 and add up to be 2.
15 and -13 since 15(-13)=-195 and 15+(-13)=2
So the factored form is
(n+15)(n-13)=0
This means we have n+15=0 and n-13=0 to solve.
n+15=0
Subtract 15 on both sides:
n=-15
n-13=0
Add 13 on both sides:
n=13
So if n=13 , then n+2=15.
If n=-15, then n+2=-13.
Let's check both results
(n,n+2)=(13,15)
13^2+15^2=169+225=394. So (13,15) looks good!
(n,n+2)=(-15,-13)
(-15)^2+(-13)^2=225+169=394. So (-15,-13) looks good!
Answer:
if u look it up it shows u on yt if that does not help im sorry
Step-by-step explanation:
Answer:
AB = 10
Step-by-step explanation:
Use the distance formula: √(12 - 18)^2 + (1 - 9)^2
√36 + 64
10
