617513.4668 is the answer
It will be negative 2,cause your equation is already in general form but we can't have a negative on a, so we should changed it on their opposite sign, so 2 will be negative
Answer:
a. Class width=4
b.
Class midpoints
46.5
50.5
54.5
58.5
62.5
66.5
70.5
c.
Class boundaries
44.5-48.5
48.5-52.5
52.5-56.5
57.5-60.5
60.5-64.5
64.5-68.5
68.5-72.5
Step-by-step explanation:
There are total 7 classes in the given frequency distribution. By arranging the frequency distribution into the refine form we get,
Class
Interval frequency
45-48 1
49-52 3
53-56 5
57-60 11
61-64 7
65-68 7
69-72 1
a)
Class width is calculated by taking difference of consecutive two upper class limits or two lower class limits.
Class width=49-45=4
b)
The midpoints of each class is calculated by taking average of upper class limit and lower class limit for each class.

Class
Interval Midpoints
45-48 
49-52 
53-56 
57-60 
61-64 
65-68 
69-72 
c)
Class boundaries are calculated by subtracting 0.5 from the lower class limit and adding 0.5 to the upper class interval.
Class
Interval Class boundary
45-48 44.5-48.5
49-52 48.5-52.5
53-56 52.5-56.5
57-60 56.5-60.5
61-64 60.5-64.5
65-68 64.5-68.5
69-72 68.5-72.5
Answer:

Step-by-step explanation:
Incomplete question
The answer is C
The length of a,b is 15
The length of j,m is 10
The width of d,c is 5
The width of m,l is 6
Length over length does not equal
Width over width