There are 64 subsets, and they are:{}, {A}, {1}, {2}, {3}, {4}, {5}, {A,1}, {A,2}, {A,3}, {A,4}, {A,5}, {1,2}, {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3, 5}, {4,5}, {A, 1, 2}, {A, 1, 3}, {A, 1, 4}, {A, 1, 5}, {A, 2, 3}, {A, 2, 4}, {A, 2, 5}, {A, 3, 4}, {A, 3, 5}, {A, 4, 5}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}, {A, 1, 2, 3}, {A, 1, 2, 4}, {A, 1, 2, 5}, {A, 1, 3, 4}, {A, 1, 3, 5}, {A, 1, 4, 5}, {A, 2, 3, 4}, {A, 2, 3, 5}, {A, 2, 4, 5}, {A, 3, 4, 5}, {1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 4, 5}, {1, 3, 4, 5}, {2, 3, 4, 5}, {A, 1, 2, 3, 4}, {A, 1, 2, 3, 5}, {A, 1, 2, 4, 5}, {A, 1, 3, 4, 5}, {A, 2, 3, 4, 5}, {1, 2, 3, 4, 5} {A, 1, 2, 3,,4, 5} .
1 4 5 1 3 6 2 4 4 6 3 1 8 1 1 3 3 4 1 2 7 2 3 5 4 1 5 5 1 4 1 5 4 6 3 1 1 1 8 7 2 1 4 3 3 3 4 3 5 3 2 2 5 3 5 4 1 6 1 3 1 6 3 2 1 7 1 7 2 1 8 1 4 4 2 4 2 4 2 7 1 4 5 1
2 and 275/300 which simplifies to 2 11/12, or 2.916666..... this is possible as ((6*1)+(9*2)+(8*3)+(8*4)+(5*5))/(36).
3 × 2 × 5 × 1 × 2 × 3 = 180
1+1=2 : 1+2=3 : 2+3=5 : 3+5=8Therefore the next number is 8
5/(√3 - 1)= 5(√3 + 1)/(√3 - 1)(√3 + 1)= (5√3 + 5)/[(√3)2 - 12)= (5√3 + 5)/(3 - 1)= 5√3 + 5)/2= 5√3/2 + 1/2
3 ÷ 2/5 = 3/1 ÷ 2/5 = 3/1 × 5/2 = (3×5)/(1×2) = 15/2 = 7½
2==5 and 5!=3 or (5-2)^2>=1
Who: DixieLocation: Hotton, Desert IsleKeystrokes:1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 8, 6,7, 5, 6, 4, 5, 3, 4, 2, 3, 1, 1Who: ElainaLocation: Wington, Bird IsleKeystrokes:3, 2, 1, 2, 3, 3, 3, 2, 2, 2, 3, 5, 5, 3,2, 1, 2, 3, 3, 3, 3, 2, 2, 3, 2, 1Who: HeinrichLocation: Chillton, Snow IsleKeystrokes:5, 3, 2, 1, 2, 3, 5, 3, 2, 1, 2, 3, 5, 3,5, 6, 3, 6, 5, 3, 2, 1Who: InaraLocation: Appleton, Horse IsleKeystrokes:3, 3, 4, 5, 5, 4, 3, 2, 1, 1, 2, 3, 3, 2, 2,3, 3, 4, 5, 5, 4, 3, 2, 1, 1, 2, 3, 2, 1, 1Who: JasperLocation: Earton, Horse IsleKeystrokes:1, 2, 3, 4, 5, 6, 7, 8Who: KyleighLocation: Witherton, Rider IsleKeystrokes:3, 3, 5, 3, 2, 3, 1, 5, 5, 6, 5, 4, 5, 2, 8,7, 6, 5, 4, 3, 2, 5, 3, 2, 3, 1Who: SandraLocation: Flipperton, Dolphin IsleKeystrokes:1, 1, 2, 1, 2, 3, 2, 3, 4, 3, 4, 5, 4, 5, 6, 5,6, 7, 6, 7, 8, 5, 3, 1Who: SorenLocation: Crystalton, Ice IsleKeystrokes:3, 5, 3, 2, 1, 3, 5, 3, 2, 1, 2, 3, 4, 5,6, 5, 4, 3, 2Who: VeronicaLocation: Shellton, Turtle IsleKeystrokes:3, 2, 1, 3, 2, 1, 5, 4, 4, 3, 5, 4, 4, 3, 5, 8, 8,7, 6, 7, 8, 5, 5, 5, 8, 8, 7, 6,7 8, 5, 5, 4, 3, 2, 1Who: YancyLocation: Treeton, Horse IsleKeystrokes:1, 1, 5, 5, 6, 6, 5, 4, 4, 3, 3, 2, 2, 1, 5, 5, 4, 4, 3, 3,2, 5, 5, 4, 4, 3, 3, 2, 1, 1, 5, 5, 6, 6, 5, 4, 4, 3, 3, 2, 2, 1
1 2/3=5/3 (2/3)/(5/3)=(2/3)*(3/5)=2/5
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)}
3 1/2
There are 32 possible subset from the set {1, 2, 3, 4, 5}, ranging from 0 elements (the empty set) to 5 elements (the whole set): 0 elements: {} 1 element: {1}, {2}, {3}, {4}, {5} 2 elements: {1, 2}, {1, 3}, {1, 4}, {1, 5}, {2, 3}, {2, 4}, {2, 5}, {3, 4,}, {3, 5}, {4, 5} 3 elements: {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5} 4 elements: {1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 4, 5}, {1, 3, 4, 5}, {2, 3, 4, 5} 5 elements: {1, 2, 3, 4, 5} The number of sets in each row above is each successive column from row 5 of Pascal's triangle. This can be calculated using the nCr formula where n = 5 and r is the number of elements (r = 0, 1, ..., 5). The total number of subset is given by the sum of row 5 of Pascal's triangle which is given by the formula 2^row, which is this case is 2^5 = 32.
The answer is 8 because 1+1=2, 1+2=3, 2+3=5,and 3+5=8.
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
It is Mathematically impossible. In order to find a single 7 in 1 1 1 1 2 1 1 1 3 1 2 2 2 2 3 2 2 2 The order would be as a requirement: 1 1 1 1 2 1 1 1 1 3 2 2 2 2 3 2 2 2 2 4 3 3 3 3 4 3 3 3 3 5 4 4 4 4 5 4 4 4 4 6 --- --- 5 5 5 5 6 5 5 5 5 7 (THERE!) If in that order the first 7 would be the 50th number. as from that you can see that the first 14 would be the 100th number, as well as the 21 being the 150th and so on.
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