Mean: 37.2
Median: 20
Mode: 20, 12, 61, 74, 19
Mode: 12 and 23Mean = 17.5Median = 17.5Mode: 12 and 23Mean = 17.5Median = 17.5Mode: 12 and 23Mean = 17.5Median = 17.5Mode: 12 and 23Mean = 17.5Median = 17.5
Mean: 20.2222222 Median: 12 Mode: 11 Range: 79
The mean of the numbers is 22. The median of the numbers is 21. The mode is 16, 20 and 30 or you can say the set has no mode.
Mean = 14 Median = 12 Mode = 12
mean = 19 median = 12 mode = 12
Mean: 50.6 Median: 42 Range: 85 Mode: None (all numbers occur with same frequency)
Mean - 15 ( Add them all together and divide by how many #'s you have. ) Median - 14.5 No mode (mode is number that appears the most)
6, 6, 15, 16, 17
There is no single answer to that. You could come up with many sets of numbers that would have those properties.
Mean: 11.6 Median: 12 Mode: 10, 12, 15, 16, 5 Range: 11
Mean: 17 Median: 19 Mode: 20
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
what i have is mean is 22.2,median which is the middle is 23,range is 26, and mode is 12
Mean: 11 Median: 11 Mode: 4 Range: 18
You add all the numbers up and then divide by the amount of numbers there are. e.g 2+4+6= 12, there are 3 numbers (2,4,6) therefore you divide 12 by 3, and you have 4, therefore 12 is the mean. The mode is the most common number, so if you have 3,3,4,5,5,5,5,6 and you need to find the mode, the mode is 5. The median is the middle number, so if you have 1,6,9,114,300 then 9 is the median as it is right in the middle.
Mean = 28626Median = 11.5 There is no mode.
Answer: 5,6,7,7,8,10,11,13,14,15,16,18
# 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.
Range: 34 Mean: 21.71 Median: 15 Mode: 10
Mean = 11 Median = 11 Mode = 4 Range = 18
Mean is the average of a group of numbers. To find the mean add all the numbers up then divide the result by the number of numbers. e.g. 3,6,9, and 12, you would add all those numbers up:3+6+9+12=3030/4=7.5 the answer!Median is the number that shows up the most in a group of numbers. so say you have:3,5,7,8,10,14,5,7,5there are three 5's so 5 is the median:3,5,7,8,10,14,5,7,5Mode is the "middle number" what you do is you order the numbers from least to greatest then find the number in the middle:e.g. 2,5,4,7,8,5,62,4,5,5,6,7,8 5 is the middle number so that is the answer!if their are two middle numbers, you just find the average of the two numbers.Median i when you have a series of numbers and put them in the right order for example 4,1,7,8,5 would then become 1,4,5,7,8 and then you pick the middle number which is 5 and that s the answer
If the minimum and maximum are both 3 then all 12 numbers must be 3. In that case, the median and mode must also be 3.
One possible set is {9, 9, 10, 11, 12, 15, 18}
12, 14, 15, 16, 16, 17 The median is 15.5 The mode is 16 The range is 5 The mean is 15
Make up a data set of at least 12 numbers that have the following landmarks maxium:18 range 13 mode 7 median 12