Proof.
For two dice, the number of favorable cases is 5: (6, 2);(5, 3);(4, 4);(3, 5);(2, 6); so the
probability is 5/36 =
30/216
For three dice, the number of favorable cases is 21: (5, 2, 1), (4, 3, 1) counted with all permutations
(so 6 2 = 12), and (1, 1, 6), (2, 2, 4), (3, 3, 2), counted with the distinct permutations (so
3 3 = 9). The probability is 21/216
So the more likely case is for two dice.
Answer:
81x^8
Step-by-step explanation:
i hope this helps :)
There are 3 movies. The total to rent all 3 is $13.20. To find the price of each movie, divide the total amount by 3.
Let M = amount per movie
M = (13.20)/3
M = $4.40
Answer:
![\sqrt{203}](https://tex.z-dn.net/?f=%5Csqrt%7B203%7D)
Step-by-step explanation:
1. ![11^{2} + b^{2} = 18^{2}](https://tex.z-dn.net/?f=11%5E%7B2%7D%20%2B%20b%5E%7B2%7D%20%3D%2018%5E%7B2%7D)
2. ![b^{2} =203](https://tex.z-dn.net/?f=b%5E%7B2%7D%20%3D203)
3. b = ![\sqrt{203}](https://tex.z-dn.net/?f=%5Csqrt%7B203%7D)
Can't be simplified.
Upen saves the most of his salary every month