The modes are 2, 3, 4, 5 and 6.
5/6 + 3/2 =14/6 or7/3=5/6 +3/2=5/6 +9/6= 14/6 or 7/3
2/3+5/6=4/6+5/6=9/6=3/2
Without a sign, I'm not sure what you want to have happen to them. 1/2 = 3/6 3/6 + 5/6 = 8/6 = 4/3 or 1 and 1/3 3/6 - 5/6 = -2/6 = -1/3 1/2 x 5/6 = 5/12 1/2 divided by 5/6 = 1/2 x 6/5 = 6/10 = 3/5
6 = 2×3 → 6² = (2×3)² = 2²×3² > 3² 5³ = 125, 6² = 36 Therefore 5³ is the greatest with 6² next and 3² the least (5³ > 6² > 3²).
12 - (1/6 + 2/3) = 12 - (1/6 + 4/6) = 12 - 5/6 = 11 1/6 or did you mean: 1/6 + (12 - 2/3) = 1/6 + 11 1/3 = 11 + (1/6 + 1/3) = 11 + (1/6 + 2/6) = 11 + 3/6 = 11 + ½ = 11½
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.
5/6 + 3/2 =14/6 or7/3=5/6 +3/2=5/6 +9/6= 14/6 or 7/3
Midpoint of (3, -6) and (-5, 2) = [(3-5)/2, (-6+2)/2] = (-1, -2)
2/3 + 5/6 = 4/6 + 5/6 = 9/6 =3/2
2/3+5/6=4/6+5/6=9/6=3/2
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
Without a sign, I'm not sure what you want to have happen to them. 1/2 = 3/6 3/6 + 5/6 = 8/6 = 4/3 or 1 and 1/3 3/6 - 5/6 = -2/6 = -1/3 1/2 x 5/6 = 5/12 1/2 divided by 5/6 = 1/2 x 6/5 = 6/10 = 3/5
6 = 2×3 → 6² = (2×3)² = 2²×3² > 3² 5³ = 125, 6² = 36 Therefore 5³ is the greatest with 6² next and 3² the least (5³ > 6² > 3²).
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
6! 3!2! 5!