Hello <u>Queengirl0 </u>
Past tense:
Melissa has a full chocolate bar 8/8
At lunch:
Melissa ate 2/8 of the candy bar, 8/8 - 2/8 = 6/8
After school:
Melissa ate 3/8 of the candy bar, 6/8 - 3/8 = 3/8
<u><em>Melissa has 3/8 of a chocolate bar left</em></u>
:)
Answer:
=0.1587 or 15.87%
So option A is correct answer so 15.87% of the invoices were paid within 15 days of receipt.
Step-by-step explanation:
In order to find the percent of the invoices paid within 5 days of receipt we have to find the value of Z first.
where:
X is the random varable which in our case is 15 days
u is the mean or average value which is 20 days
S is the standard deviation which is 5 days
Z=-1.0
We have to find Probability at Z less than -1
P(Z<-1.0) which can be written as:
=1-P(Z>1.0)
From Cumulative distribution table:
=1-(0.3413+0.5)
=0.1587 or 15.87%
So option A is correct answer so 15.87% of the invoices were paid within 15 days of receipt.
Answer:
80
Step-by-step explanation:
16:20*100 =
(16*100):20 =
1600:20 = 80
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