Answer:
a = 3, b = 8, c = 14
Step-by-step explanation:
First, we can organize this, putting lower values first.
2, 3, 3, 5, 9, 9, 11, 12
a , b , c somewhere there
Given this information, one thing we can look at first is the mode. The mode is 3, so there must be more 3s than any other number. Currently, there are 2 3s and 2 9s, so there must be at least one more 3 and no more 9s to make that true. Therefore, a, b, or c is 3. Therefore, we have
2, 3, 3, 3, 5, 9, 9, 11, 12
2 of a, b, c somewhere in there
Next, the median is 8. In our current state, the median is 5. There are 9 numbers, with 11 total including the 2 remaining values. Because there will be an odd amount of values, the median must be a number on the list. Therefore, our list is
2, 3, 3, 3, 5, 8, 9, 9, 11, 12
1 of a, b, c somewhere in there
There are 4 numbers above the median and 5 numbers below right now. To balance this out, there must be another number above the median. As a consequence, the remaining value must be greater than 8.
Finally, we know that the range is 12, so maximum - minimum = 12. Because the remaining number must be greater than 8, the minimum number is 2, no matter what. Therefore,
maximum - 2 = 12
add 2 to both sides to isolate maximum
maximum = 14
There is no 14 currently on the list, so the remaining value must be 14.
Our a, b, and c are as follows, in order from smallest to largest:
3, 8, 14
-1
-1
-1
-1
×
(denominator should be -2=common)
