A list contains 7 consecutive even numbers in ascending order. The largest number is 30 and the smallest is 12 less than this. The mode is 20 and median is 22.If the sum of the two remaining numbers is 50,find all the number

18, 20, 20, 22, 24, 26, 30

Step-by-step explanation:

We know that the largest number of the list is 30 and the smallest number is 12 less than that, which is 30 - 12 = 18.

This is our list so far: 18, __ , __ , __ , __ , __ , 30

The median of a set is the middle number of the set. In this case, since there are 7 terms in the set, the median, which the problem states is 22, must be the 4th number, since the 4th number is in the middle.

This is our list now: 18, __ , __ , 22, __ , __ , 30

Now, let's deal with the mode. The mode of a set is the term that shows up the most. Since the set is in ascending order and 20 < 22, we know that the two numbers in between 18 and 22 must be 20 and 20. This means that no term can show up more than once, with the exception of 20 because it is the mode.

This is our list now: 18, 20, 20, 22, __ , __ , 30

Finally, we'll deal with the last two numbers. We know that they must be even and that they both must be in between 22 and 30 (since the set is in ascending order). We also know that the two numbers must be different (see last paragraph) and that they sum to 50. The only two numbers that satisfy all of these conditions are 24 and 26, so they must be our last two numbers.

Therefore, the answer is 18, 20, 20, 22, 24, 26, 30. Hope this helps!