151
The term "outlier" is subjective. Personally, I don't see any outliers in this list.
Two numbers with a sum of 47 are 1 and 46, 2 and 45, 3 and 44, 4 and 43, etc.
The outlier to tht would be 78 because it is higher than the other numbers!
49C6 = 49!/[(49 - 6)!6!] = 49!/(43!6!) = (49*48*47*46*45*44*43!)/(43!*6*5*4*3*2*1) = 49*47*46*3*44 = 13,983,816 or 49C6 = 49*48*47*46*45*44/6*5*4*3*2*1 = 13,983,816
the range in math is to subtract the highest to the lowest and that wold be you answer like this: 44 45 46 47 48 49 50 take 50-44=6 range =6.
The term "outlier" is subjective. Personally, I don't see any outliers in this list.
40: 2, 5 44: 2, 11 45: 3, 5 47: 47
47 2/3
41: 1, 41 42: 1, 2, 3, 6, 7, 14, 21, 42 43: 1, 43 44: 1, 2, 4, 11, 22, 44 45: 1, 3, 5, 9, 15, 45 46: 1, 2, 23, 46 47: 1, 47 41: 41 42: 2, 3, 7 43: 43 44: 2, 11 45: 3, 5 46: 2, 23 47: 47
55.5
Two numbers with a sum of 47 are 1 and 46, 2 and 45, 3 and 44, 4 and 43, etc.
The outlier to tht would be 78 because it is higher than the other numbers!
43
Their product.
41,43 and 47. Since the factors are 1 and the number itself.
45 + 47 = 92
49C6 = 49!/[(49 - 6)!6!] = 49!/(43!6!) = (49*48*47*46*45*44*43!)/(43!*6*5*4*3*2*1) = 49*47*46*3*44 = 13,983,816 or 49C6 = 49*48*47*46*45*44/6*5*4*3*2*1 = 13,983,816