5/30 + 4/6 = 1/6 + 4/6 = 5/6
4/6, 5/5, 5/6, 6/4, 6/5, 6/6 ie 6 out of 36.
what is the largest three-digit number that is divisible by 4, 5, and 6
(1, 5, 6, 6), (2, 4, 6, 6), (2, 5, 5, 6), (3, 3, 6, 6), (3, 4, 5, 6), (3, 5, 5, 5), (4, 4, 4, 6), (4, 4, 5, 5).
6x4+5
infinity because numbers are endless
45
There are 216 permutations of three standard dice. Of these, there are exactly 10 that sum to 15, namely 3-3-6, 4-5-6, 4-6-5, 5-4-6, 5-5-5, 5-6-4, 6-3-6, 6-4-5, 6-5-4, and 6-6-3. The probability, then, of rolling a sum of 15 with three standard dice is 10 in 216, or 5 in 108, or about 0.04630.
5/30 + 4/6 = 1/6 + 4/6 = 5/6
Hey ppl sorrry it's not necessary anymore and the answer is not 39.
5/30 + 4/6 = 1/6 + 4/6 = 5/6
4/6, 5/5, 5/6, 6/4, 6/5, 6/6 ie 6 out of 36.
-6, -5 and -4.
what is the largest three-digit number that is divisible by 4, 5, and 6
(1, 5, 6, 6), (2, 4, 6, 6), (2, 5, 5, 6), (3, 3, 6, 6), (3, 4, 5, 6), (3, 5, 5, 5), (4, 4, 4, 6), (4, 4, 5, 5).
The sum of 5/6 and 4/9 is 1 5/18
On two 6-sided fair dice:Rolling a sum of 5: 4 out of 36, or 11.11%Rolling a sum of 7: 6 out of 36, or 16.67%Rolling a sum of either 5 or 7: 10 out of 36, or 27.78%Here's why: Look at the dice as unique, (maybe one is red and one is blue). Then there are 36 different possible outcomes.The four that give a sum of 5: 1+4, 2+3, 3+2, 4+1.The six that give sum of 7: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1So there are 4 chances out of 36 (or 1 out of 9 = 11.11%) to get a sum of 5, and 6 chances out of 36 (or 1 out of 6 = 16.67%) to get a sum of 7. If you want either 5 or 7 as a sum, add the chances together to get 10 out of 36 (or 5 out of 18 = 27.78%)