mean: 9 ordered sample: 7, 8, 8, 10, 12 median: 8 mode: 8 range: 12 - 7 = 5
5, 5, 8, 10, 12.
Oh, dude, let me break it down for you. If the mode is 8, that means one of the numbers has to be 8 since it's the most frequent one. To get a mean of 10, the other two numbers need to balance out the total. So, we can have 8, 10, and 12. Ta-da!
1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 6, 8, 10, 12 The mean is 4 The median is 2.5 The mode is 1
8, 8, 8, 10, 14, 15 The mean is 10.5 The median is 9. The mode is 8. The range is 7.
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
{5, 5, 8, 8, 12}{5, 5, 8, 8, 12}{5, 5, 8, 8, 12}{5, 5, 8, 8, 12}
mean = 19 median = 12 mode = 12
Mean: 11.6 Median: 12 Mode: 10, 12, 15, 16, 5 Range: 11
8+10+12=30 30/3=10 10 is the average of 8 12 and 10 If that's 812 and 10, the average is 411. If that's 8, 12 and 10, the average is 10.
8+10+12=30 30/3=10 10 is the average of 8 12 and 10 If that's 812 and 10, the average is 411. If that's 8, 12 and 10, the average is 10.
The modes is the number which occurs most often. The range is the biggest number - the smallest number. The only way to have a mode of 10 and a range of 8, is to have the two numbers be 10 and 2. The numbers are different so each is a mode and the range is 10-2-8. Note, 2 is also a mode in the set {2,10}.
The greatest common factor of 8, 10, and 12 is 2.
8+10+12=30 30/3=10 10 is the average of 8 12 and 10 If that's 812 and 10, the average is 411. If that's 8, 12 and 10, the average is 10.
it has manager mode 8 v 8 custom tourneys and online mode Thanks, but what is 8 v 8 custom tourneys?
So first let's order the set: 4,8,12,12,15,16 The mode is 12, there are 2 of them and 1 of everything else. The median is also 12 and the mean is (4+8+12+12+15+16)/6
12 and 10/8 = 13 and 2/8 = 131/4