To solve the expression (9 + 2) x 6 - 5, first, calculate the sum inside the parentheses: 9 + 2 = 11. Then, multiply by 6: 11 x 6 = 66. Finally, subtract 5: 66 - 5 = 61. Therefore, the result is 61.
5,6,2,3,9 has a greater sum than 10,4,5,1 and 2 which has a sum of 22.
When 2 dice are rolled, the probability of getting a sum less than 4 or greater than 9 is 2 in 9. There are 36 permutations of two six sided dice, 1-1, 1-2, ..., 1-6, 2-1, ... 2-6, ..., 6-1, 6-2, ... 6-6. Of these permutations, 3 have a sum less than four, 1-1, 1-2, and 2-1, and 5 have a sum greater than nine, 4-6, 5-5, 5-6, 6-4, 6-5, and 6-6. Add these probabilities together and you get (3 + 5) in 36, or 8 in 36, or 2 in 9.
43
A sum of 9 with 2 dice can be achieved with 3+6, 4+5, 5+4, and 6+3, so there are 4 elements in the event space for a sum of 9. This corresponds to a probability of 4 in 36, or 1 in 9, or about 0.1111 for that result.
There are 36 permutations of two dice. Of those, 9 add up to a sum of 5 or 6... 1 + 4 = 5 2 + 3 = 5 3 + 2 = 5 4 + 1 = 5 1 + 5 = 6 2 + 4 = 6 3 + 3 = 6 4 + 2 = 6 5 + 1 = 6 The probability, then, of rolling a 5 or 6 with two dice is 9 in 36, or 1 in 4, or 0.25.
There is no sequence of adds or subtracts of 5, 6, 8, 9, and 10 that sum to 1. Check it with this C++ code... for (int i=0; i<32; ++i) { int sum = 0; if (i&1) sum += 5; else sum -= 5; if (i&2) sum += 6; else sum -= 6; if (i&4) sum += 8; else sum -= 8; if (i&8) sum += 9; else sum -= 9; if (i&16) sum += 10; else sum -= 10; cout << i << " " << sum << endl; }
6+5=11 2 ess gives you 9
5,6,2,3,9 has a greater sum than 10,4,5,1 and 2 which has a sum of 22.
When 2 dice are rolled, the probability of getting a sum less than 4 or greater than 9 is 2 in 9. There are 36 permutations of two six sided dice, 1-1, 1-2, ..., 1-6, 2-1, ... 2-6, ..., 6-1, 6-2, ... 6-6. Of these permutations, 3 have a sum less than four, 1-1, 1-2, and 2-1, and 5 have a sum greater than nine, 4-6, 5-5, 5-6, 6-4, 6-5, and 6-6. Add these probabilities together and you get (3 + 5) in 36, or 8 in 36, or 2 in 9.
43
There are 36 permutations of two dice. Of them, 6 have a sum of 7, (1+6, 2+5, 3+4, 4+3, 5+2, and 6+1) and 2 have a sum of 11 (5+6 and 6+5). The probability, then of getting a sum of 7 or 11 is (6 plus 2) in 36, or 8 in 36, or 2 in 9, or about 0.2222.
The sum of 6/18 and 2/9 is 5/9.
4/6, 5/5, 5/6, 6/4, 6/5, 6/6 ie 6 out of 36.
A sum of 9 with 2 dice can be achieved with 3+6, 4+5, 5+4, and 6+3, so there are 4 elements in the event space for a sum of 9. This corresponds to a probability of 4 in 36, or 1 in 9, or about 0.1111 for that result.
6(5 + 9) = (6 x 5) + (6 x 9) = 6 x 14 = 84
There are 36 permutations of two dice. Of those, 9 add up to a sum of 5 or 6... 1 + 4 = 5 2 + 3 = 5 3 + 2 = 5 4 + 1 = 5 1 + 5 = 6 2 + 4 = 6 3 + 3 = 6 4 + 2 = 6 5 + 1 = 6 The probability, then, of rolling a 5 or 6 with two dice is 9 in 36, or 1 in 4, or 0.25.
To determine if 455 is divisible by 2, 3, 5, 6, or 9, we can check each divisor. 455 is odd, so it is not divisible by 2. The sum of the digits (4 + 5 + 5 = 14) is not divisible by 3, so 455 is not divisible by 3 or 6. It ends in 5, indicating it is divisible by 5, but not by 9 as the sum of the digits is not a multiple of 9. Therefore, 455 is only divisible by 5.