What is the interquartile range of 24,36,42,57,65

Hello!

 First arrange the numbers in ascending order. The numbers are already in ascending order. 

The IQR of a distribution is the difference between the upper and lower quartiles. 

Find the median first. The median is the middle number when arranged in ascending or descending order, which is, in this case, 42. 

Now find the upper and lower quartiles. These are the medians of the lower and upper halves of data. 42 divides the data set into two halves... 

24, 36 --- 42 --- 57, 65 

Since there are two numbers in each half, you can use the median formula (n + 1) / 2 to get (2 + 1)/2 = 3/2 = 1.5. That means the median is the 1.5th number. There is no 1.5th number in the lower half (or the upper half). The first number is 24, and the second number is 36. So, the 1.5th number is the mean of these two numbers, or the number that is exactly halfway between 24 and 36 (similarly for the upper quartile). 

Lower quartile (Q1) = (24 + 36) / 2 = 60/2 = 30 
Upper quartile (Q3) = (57 + 65) / 2 = 122/2 = 61 

The interquartile range (IQR) is the difference between the two quartiles, Q1 and Q3. 

So, IQR = Q3 - Q1 = 61 - 30 = 31. 

I Hope my answer has come to your Help. Thank you for posting your question here in . We hope to answer more of your questions and inquiries soon. Have a nice day ahead! :)

(Mark As Brainliest IF Helped!)

-TheOneAboveAll :D

The answer is C : The area of sector AOB is equal to the area of sector DOC because the central angle measures are equal.

Step-by-step explanation:

I just took the test