There are an even number of numbers so you have to take the middle two and get the average of them. The middle 2 numbers are 108 and 110. Add them together and divide by 2 and you get 109.
mean: 104.4 median: 104 mode: 104 range: 5
103 is the only one.
106 is the median, as it is the number halfway between the two values, or half of the total of the two numbers.
3
(0/101) + (0/102) + (1/103) + (0/104) + (0/105) + (1/106)
mean: 104.4 median: 104 mode: 104 range: 5
103 is the only one.
106 is the median, as it is the number halfway between the two values, or half of the total of the two numbers.
3
Ruthenium isotopes are: Ru-96, 98, 99, 100, 101, 102, 103, 104, 106.
The median of the numbers is equal to the number in the middle when all are arranged in numerical order. In this case, the median number is 99.
(0/101) + (0/102) + (1/103) + (0/104) + (0/105) + (1/106)
No, it does not.
There is an infinite number of possibilities: ... , 101+4.7, 102+3.7, 103+2.7, 104+1.7, 105+0.7, 106+(-0.3), 107+(-1.3), ... ..., 101.1+4.6, 102.1+3.6, ... ..., 101.15+4.55, ... etc. And all that is before bringing in irrational numbers.
1 undecillion and one,1 undecillion and two,1 undecillion and three,etc.Or do you mean in terms of the next power of a million (on the long scale) or thousand (on the long scale or plus one on the short scale)?Long scale, as used in countries like Europe: continues:undecillion = (106)11 = 1066 1069 = 103 x (106)11 = undecilliard1072 = (106)12 = duodecillion1075 = 103 x (106)12 = duodecilliard1078 = (106)13 = tredecillion1081 = 103 x (106)13 = tredecilliardetcShort scale as used in countries like USA, continues:undecillion = (103)11+1 = 1036 1039 = (103)12+1 = duodecillion1042 = (103)13+1 = tredecillionetc
Your question was a little unclear in two ways: You said "using 3 digits". Are 1 and 2 digit numbers allowed? and If you reorder the numbers in a set, do you count it twice? For instance, if you had already counted the set 101, 102, 103, 104, 105, 106, would you also count 106, 105, 104, 103, 102, 101? If yes to both questions, there are 985,084,775,273,880,000 different sets. If yes to the first question, no to the second, there are 1,368,173,298,991,500 different sets. If no to the first question, yes to the second, there are 522,639,254,696,832,000 different sets. If no to both questions, there are 725,887,853,745,600 different sets. Hope this helped :).
20,965,304.405 = (2 x 107) + (0 x 106) + (9 x 105) + (6 x 104) + (5 x 103) + (3 x 102) + (0 x 101) + (4 x 100) + (4/101) + (0/102) + (5/103)