Janelle made ordered lists of the heights, in inches, of the students in ms. carver’s and mr. wright’s homerooms. she found the two lists had the same median value, but the list from ms. carver’s class showed greater variability. if the shortest student in mr. wright’s homeroom is 48 inches tall and the tallest student is 64 inches tall, what could be the heights of the shortest and tallest students in ms. carver’s homeroom?

I will assume we are working with normal distributions.

The fact that the Ms. Carver's homeroom has greater variability can imply that there are more "diversity" in heights.

If the minimal and maximal in Mr wright's are 48 inches and 64 inches, then the minimal in ms. carver's must be less than 48 inches, and the maximal must be greater than 64 inches (notice that the change in the minimal and maximal must be the same in modulus, because the media must stay the same).

For example, the minimal can be 46 inches and the maximal 66 inches, wich will induce a bigger variance than the other class.

yes, gavin is right.

step-by-step explanation:

gavin is correct. 

this problem is about converting 9% per year to how many percent per quarter. the rate is still the same. it only differed in the time period. so, through dimensional analysis,

9% per year  × (1 year / 4 quarters) = 2.25% per quarter

if he was asked to convert 9% per year to how many percent per quarter  compounded quarterly, then you will not simply divide by four. this is because compounding is not linear.

hope this , have a good day. c;