You have to keep order of operations in mind.
1. Take care of the parentheses; distribute.
-3+6x-3=-20-8x
Simplified: 6x-6=-20-8x
2. Isolate the x variable.
Add 8x to both sides and add 6 to both sides.
14x=-14
3. Solve for x.
Divide by 14 on both sides.
x=-1
Assuming that CB is tangent, this is just a right triangle with a hypotenuse of 20 and a side of 12 so by the Pythagorean Theorem:
20^2=12^2+x^2
400=144+x^2
x^2=400-144
x^2=256
x=16
<h3>
Answer: 20% increase</h3>
======================================================
Explanation:
A = old value = 110
B = new value = 132
C = percent change
C = 100*(B-A)/A
C = 100*(132-110)/110
C = 20%
The positive C value means we have a percent increase, which corresponds to the fact that A = 110 increases to B = 132.
The B-A = 132-110 = 22 portion up top indicates that the increase is 22 pounds, which when divided over 110, gets us 22/110 = 0.20 = 20%
---------------
Checking the answer:
20% of 110 = 0.20*110 = 22
She's able to lift 22 extra pounds on top of the 110 pounds she can handle previously, so she can now lift 110+22 = 132 pounds. The answer is confirmed.
Answer:
it descended 31 Feet b b b b b b b b b bb b b b b b b b b b b
I'll leave the computation via R to you. The 
are distributed uniformly on the intervals ![[0,10i]](https://tex.z-dn.net/?f=%5B0%2C10i%5D)
, 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.
