8
Assuming by "numbers" you mean "whole numbers":{10, 10, 12, 13, 15}If the five numbers are {a, b, c, d, e} with a ≤ b ≤ c ≤ d ≤ e, then:median = 12 → c = 12leaving only two numbers (a, b) ≤ 12.mode = 10 → two or more numbers equal 10 (which is less than 12) → a = b = 10The final two numbers (d, e) are not equal, both greater than 12, and such that the sum of all five numbers is 60.10 + 10 + 12 + d + e = 60 → d + e = 28 → d = 13, e = 15→ {a, b, c, d, e} = {10, 10, 12, 13, 15}[If "numbers" includes "decimal numbers" (ie numbers with a fractional part), then as long as d + e = 28 and 12 < d < e there are infinitely many solutions, eg {10, 10, 12, 12.5, 15.5}, {10, 10, 12, 13.75, 14.25}]
That set has no mode.
All of the numbers appear once and as a mode is the most commonly occurring number, there is no mode.
The mean is the average. For example if you had the numbers 11, 10, 12, 11, 7, and 15. You would add them up and divide by how many numbers there are (which in this case is 6). The number you get is your mean (in this case the means in 11).The mode is the number that occurs most often. Using the same set of numbers above (11, 10, 12, 11, 7, and 15), the mode would be 11 because that's the only number that occurs the most. In other words, there is more than one of the same number.
# Well, the mean is like the average. So if you had these numbers 5,10,and 15, the mean(average) would be 10. # The median is like the middle number in a set of numbers. So if these were your numbers 3, 54, 66,71,75, 76, 78, 84, 88, 92, 99. The median is 76. # The mode is the number that repeated itself more then once. (there can be more than 1 mode) So if these were your set of numbers, 12, 12, 12, 43, 53, 61, 78, 84, 94, 99,99,99,99. Then your mode would be 99. # The range is the difference between the min. and max. numbers in your set of number. So if your set of numbers were 15,34,57,78,81,97. The range would be 82.
There are three modes: 11, 12 and 13
There is no single answer to that. You could come up with many sets of numbers that would have those properties.
Mean: 20.2222222 Median: 12 Mode: 11 Range: 79
Assuming by "numbers" you mean "whole numbers":{10, 10, 12, 13, 15}If the five numbers are {a, b, c, d, e} with a ≤ b ≤ c ≤ d ≤ e, then:median = 12 → c = 12leaving only two numbers (a, b) ≤ 12.mode = 10 → two or more numbers equal 10 (which is less than 12) → a = b = 10The final two numbers (d, e) are not equal, both greater than 12, and such that the sum of all five numbers is 60.10 + 10 + 12 + d + e = 60 → d + e = 28 → d = 13, e = 15→ {a, b, c, d, e} = {10, 10, 12, 13, 15}[If "numbers" includes "decimal numbers" (ie numbers with a fractional part), then as long as d + e = 28 and 12 < d < e there are infinitely many solutions, eg {10, 10, 12, 12.5, 15.5}, {10, 10, 12, 13.75, 14.25}]
5, 5, 8, 10, 12.
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!
(4, 4, 6, 10, 12)
10 is the mode, as it is the most commonly occurring number.
if the 12 digits above are 12 separate numbers the mode, or most common number is 1.
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
One possible set is {9, 9, 10, 11, 12, 15, 18}
That set has no mode.