Question

Think of 5 positive integers that have a mode of 3, a median of 6, a mean of 6 and a range of 8.

Answer 1

Answer:

The answer is given below

Step-by-step explanation:

For the mode to be 6, there has to be at least two 6s.

For the median to be 6, with two 6s, there must be at least one number above 6.

For the mean to be 6, the sum is 30, so the three remaining numbers must total 18.

Let x, y & z be the remaining numbers in order of value.

So x + y + z = 18,

and x < y < z as if any are equal, we’d have two modes

For a range of 6, z = x + 6, so we’re left with

2x + y + 6 = 18

or 2x + y = 12

or y = 12 - 2x

If x = y, then y = 12 - 2y or 3y = 12 or y = 4

so x < 4 and z < 10

so x, y , z is one of {1, 10, 7}, {2, 8, 8}, {3, 6, 9}

and the last of these is the only one with no repeats and y < z

So the answer is: 3, 6, 6, 6, 9

Answer 2

Answer :

Answer :3, 3, 6, 7, 11

Step by Step explanation:

(I didn't use any formula to solve this)

So, the numbers are 5 positive integers which means all the numbers are greater than 0.

Now we have _, _, _, _, _.

The median is given as 6. So it will be in the centre.

Now we have _, _, 6, _, _.

The mode is given as 3. Thus 3 will appear at least 2 times to be considered mode. It can't be 3 times because 6 is median.

Now we have, 3, 3, 6, _, _.

The range is given as 8

Range = Max valve - Min value

8 = Max - 3

Maximum value = 8 + 5 = 11

Now we have, 3, 3, 6, _, 11.

The mean is given as 6

So using the mean's formula (3 + 3 + 6 + x + 11)/5 = 6

(let the missing number be x)

=> (23 + x)/5 = 6

=> 23 + x = 6 × 5 = 30

=> x = 30 - 23 = 7

Finally we have 3, 3, 6, 7, 11.

You can check by solving them that they do satisfy the conditions of the question.

Mark it as brainliest if you are satisfied^^