Answer:
15/20 and 18/20
Step-by-step explanation:
Least common denomintor of 3/4 and 9/10 = least common multiple of 4 and 10
multiples of 4: 4, 8, 12, 16, 20
multiples of 10: 10, 20
Least common denominator = 20
Rewrite fractions: 15/20 and 18/20
Answer:
No solution
Step-by-step explanation:
-3x-9=-3x-3-5
-3x-9=-3x-8
-9=-8
What is a composite number between 60 and 100? The answer is 62
If you need part A, draw a person and label them 6 feet tall and a shadow coming off of him that is five feet long and draw a basketball hoop that and label it X feet tall and draw a shadow coming from it that is 8 feet long. To solve, you use proportions 6/5=X/8 solving this you get X=9.6 feet which is the height of the rim. Therefore, the rim does not meet regulation height, since the required height (10 feet) is greater than the actual height (9.6 feet).
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.