what is the mode of the following data 18, 17, 12, 14, 8, 21, 10, 11, 19, 20, 10, 5, 17, 12, 10, 20
2 and 2 fifths = 12/5. (2/5)/(12/5) = (2/5) x (5/12) = 10/60 = 1/6.
12/5 = 10 2/5
(5/6) * (1/10) + (2/3) / (4/11) = (5/6) * (1/10) + (2/3) * (11/4)= (1/12) + (11/6) = 1/12 + 22/12 = 23/12 or 111/12(5/6) * (1/10) + (2/3) / (4/11) = (5/6) * (1/10) + (2/3) * (11/4)= (1/12) + (11/6) = 1/12 + 22/12 = 23/12 or 111/12(5/6) * (1/10) + (2/3) / (4/11) = (5/6) * (1/10) + (2/3) * (11/4)= (1/12) + (11/6) = 1/12 + 22/12 = 23/12 or 111/12(5/6) * (1/10) + (2/3) / (4/11) = (5/6) * (1/10) + (2/3) * (11/4)= (1/12) + (11/6) = 1/12 + 22/12 = 23/12 or 111/12
10 appears more times than any other number, so it is the mode.
Mean: 11.6 Median: 12 Mode: 10, 12, 15, 16, 5 Range: 11
what is the mode of the following data 18, 17, 12, 14, 8, 21, 10, 11, 19, 20, 10, 5, 17, 12, 10, 20
Yes there can be more than one mode for a set of number. There can be more than 2 modes too. Say that this is your set of numbers: 1, 3, 4, 5, 5, 6, 7, 8, 10, 10, 12, and 20. The numbers that occur the most are 5 and 10 so your mode is 5 and 10. Hope that helped!
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2
5, 5, 8, 10, 12.
Mean = 7.9 Median = 7 and 8 Mode = 4 and 12 Range = 2 to 15.
(3+5/10)(2+2/5) = 35/10 * 12/5 = 7/5 * 12/5 = 84/5
mean: 9 ordered sample: 7, 8, 8, 10, 12 median: 8 mode: 8 range: 12 - 7 = 5
5/12 = 10/24 This is correct because you just multiply BOTH the numerator and denominator by 2. 5X2=10. 12X2=24. So, if you reduce 10/24, it would be 5/12. Half of 10 is 5, or 10 divided by 2 is 5. Half of 24 is 12, or 24 divided by 2 is 12.
The average, median and mode of a list of numbers will be the same when the middle (or mean of the two middle numbers) is equal to the most common number in the list, and that number is also the mean. This assumes that the list has only a single mode. If you arrange such a list of numbers from least to greatest, the mode value will be grouped at the middle of the list, thus becoming the median as well. The average of the non-median values will be equal to the median/mode. Given a target mean/median/mode and a list length, you can construct an infinite number of lists that qualify. Here are some examples: 10 10 10 10 10 (or any list of only one number) 11 12 12 12 12 13 1 2 3 4 5 5 5 6 7 8 9
I'm going to guess that's the set (5, 5, 6, 7, 8, 8, 10, 10, 10, 12) Sum: 81 Mean: 8.1 Mode: 10 Range: 7