a school has 2 computer labs.Each lab has 30 computers. A total of 6 computers in the school are not working. which experission
2 answers:
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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:
D is correct. ( 4m - 5 ) + ( -6m + 7 - 2n )
Step-by-step explanation:
We are given a difference of two polynomial
(4m-5) - (6m-7+2n)
We need to choose correct choice which is equivalent to given expression.
Here we have minus sign between two parentheses.
First we distribute minus sign over (6m-7+2n) using distributive property.
Minus sign will distribute with each term inside (6m-7+2n).
When it will distribute it changes the sign of each term inside the parentheses.
After distribution we get
( 4m - 5 ) + ( -6m + 7 - 2n )
Thus, D is correct option which is similar to given expression.