<span>23863.05 is the answer to the problem</span>
Answer:
1099/11=99.9
since you can't have 99 boxes, you round to 100 boxes.
you will need 100 boxes.
Answer: 4.1
Step-by-step explanation:
Given : An urn contains 10 balls with the number ‘5’ printed on them, 6 balls with ‘4’ printed on them, and 4 balls with ‘2’ printed on them.
Total balls = 10+6+4=20
Let A , B and C are the events of drawing ball with the number ‘5’, ball with the number ‘4’ and ball with the number ‘2’ respectively.
Then,

If X is the number observed.
Since ![E[X]=\sum_{i=1}^{n}x_ip_i](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dx_ip_i)
Then,
![E[X]=5\times P(A)+4\times P(B)+2\times P(C)\\\\=5\times 0.5+4\times 0.3+2\times 0.2\\=4.1](https://tex.z-dn.net/?f=E%5BX%5D%3D5%5Ctimes%20P%28A%29%2B4%5Ctimes%20P%28B%29%2B2%5Ctimes%20P%28C%29%5C%5C%5C%5C%3D5%5Ctimes%200.5%2B4%5Ctimes%200.3%2B2%5Ctimes%200.2%5C%5C%3D4.1)
Hence, E[X]=4.1
Answer:
1/16 my bad
Step by step explanation:
I can‘t really draw it on here but i would help otherwise