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If the set of numbers is sorted: 1 1 2 3 4 5 5, then you can see that 1 and 5 both lie at the outer limits of the range of data.
4 is even!
It's possible. An outlier is a number that affects the the mean of a group of numbers greatly. For example the mean in this set of numbers (2, 4, 1, 5) is 3, but if I add the number 93 the new answer is 21.
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} .
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
(3 1/5) - (1 3/5) : If we borrow 1 from 3, then 3 1/5 becomes 2 6/5 [1 = 5/5, and 1/5+5/5 = 6/5]. Now can subtract 3/5 from 6/5, and subtract 1 from 2:6/5 - 3/5 = 3/5. 2-1 = 1. So the answer is 1 3/5.
2==5 and 5!=3 or (5-2)^2>=1
1 5/2 - 3/2 = (5-3)/2 = 2/2 = 1
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1 2/3=5/3 (2/3)/(5/3)=(2/3)*(3/5)=2/5
2: 1 and 2 3: 1 and 3 5: 1 and 5