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The modes are 2, 3, 4, 5 and 6.
1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6As shown above, a digit is repeated to match the value of that digit.
6 1/2 - 3 3/4. Convert 1/2 to fourths =(2/4) --> 6 2/4 - 3 3/4. You can borrow one (4/4) from the 6, so 6 2/4 = 5 6/4, then you can subtract:5 6/4 - 3 3/4 = 2 3/4
If you mean 1*2 + 2*3*3*4 - 4*5*5*6 then you have 2 + 48 - 600 = -550
7!/2! = 7*6*5*4*3 = 25207!/2! = 7*6*5*4*3 = 25207!/2! = 7*6*5*4*3 = 25207!/2! = 7*6*5*4*3 = 2520
The sample space is the following set: {(1. 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2. 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3. 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4. 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5. 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6. 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
The modes are 2, 3, 4, 5 and 6.
1-1 1-2 1-3 1-4 1-5 1-6 2-1 2-2 2-3 2-4 2-5 2-6 3-1 3-2 3-3 3-4 3-5 3-6 4-1 4-2 4-3 4-4 4-5 4-6 5-1 5-2 5-3 5-4 5-4 5-6 6-1 6-2 6-3 6-4 6-5 6-6 So there ARE 36 possible outcomes, you see. Answer BY: Magda Krysnki (grade sevener) :P
The harmonica part in Electric Worry is played on a C harp. The Tabs are: Electric Worry Harp Taps -3 +4-3+4-3´-2 ->-1-2-3´+4 -3´-3´ -2-2" -2 ->-4+5-5-5 -6+6-6-6 -3+4 -3+4-4 -3+4-4 -3+4-4 -3+4-4 -4+4-3´-2 -2-2-3´+4-4-3-2-2 -2"-2-3´+4-2 -2"-2-3´+4 -5-5-5 -6-6-6-6 -5-4+5 -4-4+4 -5+4 -3-3-3-3´-3´-2...
In a combination the order does not matter, so they are: 1 1 , 1 2 , 1 3 , 1 4 , 1 5 , 1 6 2 2 , 2 3 , 2 4 , 2 5 , 2 6 3 3 , 3 4 , 3 5 , 3 6 4 4 , 4 5 , 4 6 5 5 , 5 6 6 6
If you roll two dice, the following reuslts are possible: 2: 1+1 3: 1+2 , 2+1 4: 1+3, 2+2, 3+1 5: 1+4, 2+3, 3+2, 4+1 6: 1+5, 2+4, 3+3, 4+2, 5+1 7: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1 8: 2+6, 3+5, 4+4, 5+3, 6+2 9: 3+6, 4+5, 5+4, 6+3 10: 4+6, 5+5, 6+4 11: 5+6, 6+5 12: 6+6 As you can see, the greatest number of permutations result in a total of 7. Its probability is 6/36 or 1/6.
1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6As shown above, a digit is repeated to match the value of that digit.
4* 1 4*2 4*3 4*5 5*1 5*2 5*3 5*4 5*5\ 6*16*2 6*3 6* 4 6*5
6*6=36 1&1; 1&2; 1&3; 1&4; 1&5; 1&6 2&1; 2&2; 2&3; 2&4; 2&5; 2&6 3&1; 3&2; 3&3; 3&4; 3&5; 3&6 4&1; 4&2; 4&3; 4&4; 4&5; 4&6 5&1; 5&2; 5&3; 5&4; 5&5; 5&6 6&1; 6&2; 6&3; 6&4; 6&5; 6&6 For adding up roll one and roll two, the outcomes are 1+1= 2 1+2=2+1= 3 1+3=2+2=3+1= 4 1+4=2+3=3+2=4+1= 5 1+5=2+4=3+3=4+2=5+1= 6 1+6=2+5=3+4=4+3=5+2=6+1= 7 2+6=3+5=4+4=5+3=6+2= 8 3+6=4+5=5+4=6+3= 9 4+6=5+5=6+4= 10 5+6=6+5= 11 6+6= 12 Odds of adding up to those possible outcomes. 2 = 1:36 3 = 2:36 = 1:18 4 = 3:36 = 1:12 5 = 4:36 = 1:9 6 = 5:36 7 = 6:36 = 1:6 You will total seven more times than others. 8 = 5:36 9 = 4:36 = 1:9 10= 3:36 = 1:12 11= 2:36 = 1:18 12= 1:36
It depends on what you define as an outcome. Let me take a simpler case: just two dice.1. Outcomes of possible sums of the two dice:2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 122. Outcomes of possible sets of two dice:{1, 1}, {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 2}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {3, 3}, {3, 4}, {3, 5}, {3, 6}, {4, 4}, {4, 5}, {4, 6}, {5, 5}, {5, 6}, {6, 6}3. Outcomes of possible combinations of two dice:(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)We would need to know which you need. Assuming you want all the possible combinations of ten 6-sided dice, then it is 610, or 60,466,176.
(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).
3-4, 3-4, 3-4, 4-3, 2-2 or 4-5, 4-5, 4-3, 3-3, 3-3