I'll leave the computation via R to you. The
are distributed uniformly on the intervals
, so that

each with mean/expectation
![E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i](https://tex.z-dn.net/?f=E%5BW_i%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20wf_%7BW_i%7D%28w%29%5C%2C%5Cmathrm%20dw%3D%5Cint_0%5E%7B10i%7D%5Cfrac%20w%7B10i%7D%5C%2C%5Cmathrm%20dw%3D5i)
and variance
![\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BW_i%5D%3DE%5B%28W_i-E%5BW_i%5D%29%5E2%5D%3DE%5B%7BW_i%7D%5E2%5D-E%5BW_i%5D%5E2)
We have
![E[{W_i}^2]=\displaystyle\int_{-\infty}^\infty w^2f_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac{w^2}{10i}\,\mathrm dw=\frac{100i^2}3](https://tex.z-dn.net/?f=E%5B%7BW_i%7D%5E2%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20w%5E2f_%7BW_i%7D%28w%29%5C%2C%5Cmathrm%20dw%3D%5Cint_0%5E%7B10i%7D%5Cfrac%7Bw%5E2%7D%7B10i%7D%5C%2C%5Cmathrm%20dw%3D%5Cfrac%7B100i%5E2%7D3)
so that
![\mathrm{Var}[W_i]=\dfrac{25i^2}3](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BW_i%5D%3D%5Cdfrac%7B25i%5E2%7D3)
Now,
![E[W_1+W_2+W_3]=E[W_1]+E[W_2]+E[W_3]=5+10+15=30](https://tex.z-dn.net/?f=E%5BW_1%2BW_2%2BW_3%5D%3DE%5BW_1%5D%2BE%5BW_2%5D%2BE%5BW_3%5D%3D5%2B10%2B15%3D30)
and
![\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BW_1%2BW_2%2BW_3%5D%3DE%5Cleft%5B%5Cbig%28%28W_1%2BW_2%2BW_3%29-E%5BW_1%2BW_2%2BW_3%5D%5Cbig%29%5E2%5Cright%5D)
![\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BW_1%2BW_2%2BW_3%5D%3DE%5B%28W_1%2BW_2%2BW_3%29%5E2%5D-E%5BW_1%2BW_2%2BW_3%5D%5E2)
We have

![E[(W_1+W_2+W_3)^2]](https://tex.z-dn.net/?f=E%5B%28W_1%2BW_2%2BW_3%29%5E2%5D)
![=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])](https://tex.z-dn.net/?f=%3DE%5B%7BW_1%7D%5E2%5D%2BE%5B%7BW_2%7D%5E2%5D%2BE%5B%7BW_3%7D%5E2%5D%2B2%28E%5BW_1%5DE%5BW_2%5D%2BE%5BW_1%5DE%5BW_3%5D%2BE%5BW_2%5DE%5BW_3%5D%29)
because
and
are independent when
, and so
![E[(W_1+W_2+W_3)^2]=\dfrac{100}3+\dfrac{400}3+300+2(50+75+150)=\dfrac{3050}3](https://tex.z-dn.net/?f=E%5B%28W_1%2BW_2%2BW_3%29%5E2%5D%3D%5Cdfrac%7B100%7D3%2B%5Cdfrac%7B400%7D3%2B300%2B2%2850%2B75%2B150%29%3D%5Cdfrac%7B3050%7D3)
giving a variance of
![\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BW_1%2BW_2%2BW_3%5D%3D%5Cdfrac%7B3050%7D3-30%5E2%3D%5Cdfrac%7B350%7D3)
and so the standard deviation is 
# # #
A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute
![\mathrm{Var}[W_1+W_2+W_3]](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BW_1%2BW_2%2BW_3%5D)
![=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])](https://tex.z-dn.net/?f=%3D%5Cmathrm%7BVar%7D%5BW_1%5D%2B%5Cmathrm%7BVar%7D%5BW_2%5D%2B%5Cmathrm%7BVar%7D%5BW_3%5D%2B2%28%5Cmathrm%7BCov%7D%5BW_1%2CW_2%5D%2B%5Cmathrm%7BCov%7D%5BW_1%2CW_3%5D%2B%5Cmathrm%7BCov%7D%5BW_2%2CW_3%5D%29)
and since the
are independent, each covariance is 0. Then
![\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BW_1%2BW_2%2BW_3%5D%3D%5Cmathrm%7BVar%7D%5BW_1%5D%2B%5Cmathrm%7BVar%7D%5BW_2%5D%2B%5Cmathrm%7BVar%7D%5BW_3%5D)
![\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BW_1%2BW_2%2BW_3%5D%3D%5Cdfrac%7B25%7D3%2B%5Cdfrac%7B100%7D3%2B75%3D%5Cdfrac%7B350%7D3)
and take the square root to get the standard deviation.
Answer: The answer is the yellow lines in the attached figure.
Step-by-step explanation: As shown in the attached figure, regular hexagon FGHIJK and square ABCD shares common centre on the co-ordinate plane and AB || FG.
We are to find the line across which the combined figure will reflect onto itself.
In the attached figure, we see two lines which are yellow in colour. We can easily detect that the figure will be reflected onto itself if these two lines acts as a mirror separately.
Hence these yellow lines are the required lines.
Ok to find this answer you first have to subtract the 16 from the 30 to find what "CB" is, which is 14. From there subtract 14 and 16 from 49 to find that the whole answer is 35. I think that is the answer but I'm not positive.
Answer: there is nothing here with the info i need 2 anwser this question srry.
Step-by-step explanation:
Step-by-step explanation:
