There are
possible outcomes. At the least, you can draw (1, 2, 3) to get a sum of 6, and at most (5, 6, 7) for a sum of 18. Then the PMF is
![P(W=w)=\begin{cases}\dfrac1{35}&\text{for }w\in\{6,7,17,18\}\\\\\dfrac2{35}&\text{for }w\in\{8,16\}\\\\\dfrac3{35}&\text{for }w\in\{9,15\}\\\\\dfrac4{35}&\text{for }w\in\{10,11,13,14\}\\\\\dfrac17&\text{for }w=12\end{cases}](https://tex.z-dn.net/?f=P%28W%3Dw%29%3D%5Cbegin%7Bcases%7D%5Cdfrac1%7B35%7D%26%5Ctext%7Bfor%20%7Dw%5Cin%5C%7B6%2C7%2C17%2C18%5C%7D%5C%5C%5C%5C%5Cdfrac2%7B35%7D%26%5Ctext%7Bfor%20%7Dw%5Cin%5C%7B8%2C16%5C%7D%5C%5C%5C%5C%5Cdfrac3%7B35%7D%26%5Ctext%7Bfor%20%7Dw%5Cin%5C%7B9%2C15%5C%7D%5C%5C%5C%5C%5Cdfrac4%7B35%7D%26%5Ctext%7Bfor%20%7Dw%5Cin%5C%7B10%2C11%2C13%2C14%5C%7D%5C%5C%5C%5C%5Cdfrac17%26%5Ctext%7Bfor%20%7Dw%3D12%5Cend%7Bcases%7D)
That is, consider all the possible sums and their integer partitions. For example,
6 = 1 + 2 + 3
7 = 1 + 2 + 4
8 = 1 + 2 + 5 = 1 + 3 + 4
9 = 1 + 2 + 6 = 1 + 3 + 5 = 2 + 3 + 4
and so on.
Then the expected value is
![\mathrm E[W]=\displaystyle\sum_wwP(W=w)](https://tex.z-dn.net/?f=%5Cmathrm%20E%5BW%5D%3D%5Cdisplaystyle%5Csum_wwP%28W%3Dw%29)
![\mathrm E[W]=\displaystyle\frac{(6+7+17+18)+2(8+16)+3(9+15)+4(10+11+13+14)+5(12)}{35}](https://tex.z-dn.net/?f=%5Cmathrm%20E%5BW%5D%3D%5Cdisplaystyle%5Cfrac%7B%286%2B7%2B17%2B18%29%2B2%288%2B16%29%2B3%289%2B15%29%2B4%2810%2B11%2B13%2B14%29%2B5%2812%29%7D%7B35%7D)
![\mathrm E[W]=12](https://tex.z-dn.net/?f=%5Cmathrm%20E%5BW%5D%3D12)
We can find the variance via
![\mathrm{Var}[W]=\mathrm E[W^2]-\mathrm E[W]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BW%5D%3D%5Cmathrm%20E%5BW%5E2%5D-%5Cmathrm%20E%5BW%5D%5E2)
![\mathrm E[W^2]=\displaystyle\sum_ww^2P(W=w)](https://tex.z-dn.net/?f=%5Cmathrm%20E%5BW%5E2%5D%3D%5Cdisplaystyle%5Csum_ww%5E2P%28W%3Dw%29)
![\mathrm E[W^2]=\displaystyle\frac{(6^2+7^2+17^2+18^2)+2(8^2+16^2)+3(9^2+15^2)+4(10^2+11^2+13^2+14^2)+5(12^2)}{35}](https://tex.z-dn.net/?f=%5Cmathrm%20E%5BW%5E2%5D%3D%5Cdisplaystyle%5Cfrac%7B%286%5E2%2B7%5E2%2B17%5E2%2B18%5E2%29%2B2%288%5E2%2B16%5E2%29%2B3%289%5E2%2B15%5E2%29%2B4%2810%5E2%2B11%5E2%2B13%5E2%2B14%5E2%29%2B5%2812%5E2%29%7D%7B35%7D)
![\mathrm E[W^2]=152](https://tex.z-dn.net/?f=%5Cmathrm%20E%5BW%5E2%5D%3D152)
![\implies\mathrm{Var}[W]=152-12^2=8](https://tex.z-dn.net/?f=%5Cimplies%5Cmathrm%7BVar%7D%5BW%5D%3D152-12%5E2%3D8)
All you have to do is plug in and multiply.
32i+30j?
I think thats what it is :/
Answer:
I will only write the answers becaus ethe explanation is kand of axiomatic
Step-by-step explanation:
1. 9
2. 49
3. 144
4. 100
5. 6
6. 9
7. 11
8. 7
9. 36, 49 64, 81
10. 1225
11. 3136
12. 4096
13. 17
14. 20
15. 22
16. 168100
17. 510
18. 52
Answer:
Karen had 45 m and m's candy.
Step-by-step explanation:
Let the number of m and m's candy be 'x'.
Now given:
Karen gave an equal amount of m and m's to herself and four friends.
So we can say that;
Number of people m and m's candy distributed equally = 5
Also Given:
Each person receives m and m's equivalent to the largest one digit number.
Now we know that;
Largest one digit number is 9.
So we can say that;
Each person receives m and m's = 9
We need to find number of m and m's Karen have.
Solution:
So we can say that;
Total number of m and m's Karen have is equal to Number of people m and m's candy distributed equally multiplied by number of m and m's can each person receives.
framing in equation form we get;
Total number of m and m's Karen had = ![9\times5 = 45](https://tex.z-dn.net/?f=9%5Ctimes5%20%3D%2045)
Hence Karen had 45 m and m's candy.
Answer:
A=700
Explaination:
A=2(wl+hl+hw)=2·(12·7+14·7+14·12)=700