Answer:
The center of that class, which is the sum of its largest and smallest values divided by 2 ⇒ E
Step-by-step explanation:
* Lets explain what is the class mid-point
- It is defined as the average of the upper and lower class limits
- The class midpoint is the lower class limit plus the upper class limit
divided by 2
- The easiest way to find the class mid-point is to add the upper
and lower boundary and divide your answer by two
- The lower limit for every class is the smallest value in that class.
- The upper limit for every class is the greatest value in that class
* <u><em>Lets solve the problem</em></u>
- It is not the largest value of that class minus the class width
- Its not the difference between the largest and smallest values of
that class
- It is not the difference between the largest and smallest values of
that class divided by 2
- It is the center of that class, which is the sum of its largest and
smallest values divided by 2
Answer:
Step-by-step explanation:
<u>The quotient of </u>
- 5/6÷ (-13/7) =
- 5/6×(-7/13) =
- - 35/78
