Multiply all and get 560
so v= 560
Calculate thevolume of each container and then subtract the values.
volume of container A = 13*19*25 = 6175 in^3
volume of container B = 13*13*31 = 5239 in^3
container A is greater by 6175-5239 = 936 in^3 than container B
Answer:
![v=\frac{48}{5}=9\frac{3}{5} \\](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B48%7D%7B5%7D%3D9%5Cfrac%7B3%7D%7B5%7D%20%5C%5C%20)
Step-by-step explanation:
![\frac{8}{5}=\frac{v}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B8%7D%7B5%7D%3D%5Cfrac%7Bv%7D%7B6%7D)
Swap sides:
![\frac{v}{6}=\frac{8}{5}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%7D%7B6%7D%3D%5Cfrac%7B8%7D%7B5%7D)
Multiply to both sides by 6:
![\frac{v}{6}\cdot6=\frac{8}{5}\cdot6](https://tex.z-dn.net/?f=%5Cfrac%7Bv%7D%7B6%7D%5Ccdot6%3D%5Cfrac%7B8%7D%7B5%7D%5Ccdot6)
Group like terms:
![\frac{1}{6}\cdot6v=\frac{8}{5}\cdot6](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B6%7D%5Ccdot6v%3D%5Cfrac%7B8%7D%7B5%7D%5Ccdot6)
Simplify the fraction:
![v=\frac{8}{5}\cdot6](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B8%7D%7B5%7D%5Ccdot6)
Multiply the fractions
![v=\frac{8\cdot6}{5}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B8%5Ccdot6%7D%7B5%7D)
Simplify the arithmetic:
![v=\frac{48}{5}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B48%7D%7B5%7D)
I'm guessing you intended to say
has mean
and standard deviation
, and
has means
and standard deviation
.
If
, then
has mean
![E[W]=E[X+Y]=E[X]+E[Y]=55](https://tex.z-dn.net/?f=E%5BW%5D%3DE%5BX%2BY%5D%3DE%5BX%5D%2BE%5BY%5D%3D55)
and variance
![\mathrm{Var}[W]=E[(W-E[W])^2]=E[W^2]-E[W]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BW%5D%3DE%5B%28W-E%5BW%5D%29%5E2%5D%3DE%5BW%5E2%5D-E%5BW%5D%5E2)
Given that
and
, we have
![\mathrm{Var}[X]=E[X^2]-E[X]^2\implies E[X^2]=36+25^2=661](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BX%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2%5Cimplies%20E%5BX%5E2%5D%3D36%2B25%5E2%3D661)
![\mathrm{Var}[Y]=E[Y^2]-E[Y]^2\implies E[Y^2]=16+30^2=916](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BY%5D%3DE%5BY%5E2%5D-E%5BY%5D%5E2%5Cimplies%20E%5BY%5E2%5D%3D16%2B30%5E2%3D916)
Then
![E[W^2]=E[(X+Y)^2]=E[X^2]+2E[XY]+E[Y^2]](https://tex.z-dn.net/?f=E%5BW%5E2%5D%3DE%5B%28X%2BY%29%5E2%5D%3DE%5BX%5E2%5D%2B2E%5BXY%5D%2BE%5BY%5E2%5D)
and
are independent, so
, and
![E[W^2]=E[X^2]+2E[X]E[Y]+E[Y^2]=661+2\cdot25\cdot30+916=3077](https://tex.z-dn.net/?f=E%5BW%5E2%5D%3DE%5BX%5E2%5D%2B2E%5BX%5DE%5BY%5D%2BE%5BY%5E2%5D%3D661%2B2%5Ccdot25%5Ccdot30%2B916%3D3077)
so that the variance, and hence standard deviation, are
![\mathrm{Var}[W]=3077-55^2=52](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BW%5D%3D3077-55%5E2%3D52)
![\implies\sqrt{\mathrm{Var}[W]}=\sqrt{52}=\boxed{2\sqrt{13}}](https://tex.z-dn.net/?f=%5Cimplies%5Csqrt%7B%5Cmathrm%7BVar%7D%5BW%5D%7D%3D%5Csqrt%7B52%7D%3D%5Cboxed%7B2%5Csqrt%7B13%7D%7D)
# # #
Alternatively, if you've already learned about the variance of linear combinations of random variables, that is
![\mathrm{Var}[aX+bY]=a^2\mathrm{Var}[X]+b^2\mathrm{Var}[Y]](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BaX%2BbY%5D%3Da%5E2%5Cmathrm%7BVar%7D%5BX%5D%2Bb%5E2%5Cmathrm%7BVar%7D%5BY%5D)
then the variance of
is simply the sum of the variances of
and
,
, and so the standard deviation is again
.