Question

(a) move the cubes so that each stack has the same number of cubes. then give the number of cubes in each stack. 6 (b) what is the mean of 5, 5, 7, 8, 8, and 3? (these are the numbers of cubes in the original stacks.) (c) are the values you found in parts (a) and (b) the same? why or why not? yes. but it didn’t have to turn out that way. when the stacks are made equal, the number of cubes in each stack may not be the mean of the original stacks. yes. when we make the stacks equal, the number of cubes in each stack must be the mean of the original stacks. no. but it didn’t have to turn out that way. when the stacks are made equal, the number of cubes in each stack may be the mean of the original stacks. no. when we make the stacks equal, the number of cubes in each stack cannot be

Explanation:

Part (a)

Assuming there are 3 stacks (since the mean calculation in part (b) has 6 numbers, maybe a typo, but let's assume original stacks are 5,5,7,8,8,3 – wait, no, part (a) likely has a visual, but from part (b) we know the original numbers are 5,5,7,8,8,3. Wait, no, maybe part (a) has, say, 3 stacks? Wait, no, part (b) has 6 numbers. Wait, maybe the original stacks are 5,5,7,8,8,3 (6 stacks). To make each stack equal, we find the total number of cubes: \(5 + 5 + 7 + 8 + 8 + 3 = 36\). Then divide by the number of stacks (6): \(36 \div 6 = 6\). So each stack has 6 cubes.

Part (b)

Step1: Sum the numbers

The numbers are 5, 5, 7, 8, 8, 3. Sum them: \(5 + 5 + 7 + 8 + 8 + 3 = 36\).

Step2: Divide by the number of terms

There are 6 terms. So mean is \(\frac{36}{6} = 6\).

When we make stacks equal (redistribute to have equal number per stack), this is equivalent to finding the mean (average) of the original counts. The mean is the total divided by the number of stacks, which is exactly what we do when equalizing the stacks (total cubes divided by number of stacks gives cubes per stack). So the values must be the same because equalizing stacks (finding the "fair share" per stack) is the definition of the mean.

Answer:

6

Part (c)