There are 4 possible answers and they are:
This is assuming only integers are allowed.
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Explanation:
The sequence starts off as {A,B,C,D,E}
The median is 8, which is the direct center (slot C).
So we can update the sequence to {A,B,8,D,E}
The mode is 7. The mode is the value that shows up the most frequent compared to the other values. This tells us that the mode must show up at least two times. In this case, it must show up exactly twice because of the A,B values which are smaller than the median 8 in slot C.
So we know that A = 7 and B = 7.
The sequence updates to {7,7,8,D,E} where D and E are not 7.
If either D and/or E were 7, then the list would be {7,7,7,8,x} or {7,7,7,7,8}; however, this would lead to the median not being 8. So this line of reasoning allows us to rule out neither D nor E being 7.
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Let's add up those five values, and divide by n = 5, to compute the mean
mean = (A+B+C+D+E)/5
mean = (7+7+8+D+E)/5
mean = (22+D+E)/5
Set this equal to the given mean of 9 and isolate the D+E expression
(22+D+E)/5 = 9
22+D+E = 5*9
22+D+E = 45
D+E = 45-22
D+E = 23
Whatever D and E are, they add to 23.
Assuming D and E are integers, there are a few ways to have D+E result in 23. The further restriction is that neither D nor E can be smaller than 8. Refer to the last section.
Here are all of the integer solutions to D+E = 23 where D and E are 8 or larger.
There are 8 items in that list above. We can rule out the bottom half of that list because it repeats the top half (just in reverse order). Also, note how E is smaller than D in those items. We'll assume that D < E.
So we can shorten that list to
This leads to the four possible answers of:
This is assuming that only integers are allowed. If non-whole values are considered, then we'd have infinitely many possible answers. This is because there are infinitely many solutions to D+E = 23 where D and E aren't restricted to the integer values only.
