2 is the answer to your question
This is a "rate of pay" problem. The amount earned is equal to the (rate of pay) times the (number of hours worked).
Let the income be represented by "i". Then the formula is i = ($10.91/hour)*w, where w is the number of hours worked and has the unit of measurement "hours."
Answer:
-7/4x-68
Step-by-step explanation:
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:
so im guessing the remainder of hw problems->
Step-by-step explanation:
24-7 =17 math problems left from doing some lunch
17-7= 10 math so its 10 math problems left