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kirza4 [7]
3 years ago
15

107,609 Divided By 72

Mathematics
2 answers:
Veseljchak [2.6K]3 years ago
8 0
The answer is 1494.569444
Liula [17]3 years ago
8 0
107,609/72= 1,494.569444
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Sorting through unsolicited e-mail and spam affects the productivity of office workers. An InsightExpress survey monitored offic
nydimaria [60]

Answer:

1. Refer to the explanation for the frequency table.

2. (a) 60%

   (b) 25%

Step-by-step explanation:

Part 1. The question is asking us to group the data according to the classes given and fill in the frequency (F), relative frequency (RF), cumulative frequency (CF) and relative cumulative frequency (RCF).

To compute the frequency for each class count the number of data that lies in the class range and write it. For the first class (1-5) we can see that there are 12 numbers which lie in the range 1 to 5. Those numbers are: 2, 4, 4, 1, 2, 1, 5, 5, 5, 3, 4, 4. Similarly, for the second class, the frequency is 3 because 8, 8, 8 lie in this range from our existing data. For the third class 11-15 there are 2 numbers in the given data which lie in this range and those numbers are 12, 15. Similarly, the rest of the frequencies can be computed.

For the relative frequency, the frequency of each class is divided by the total frequency i.e. 20.  

For the class 1-5, RF = 12/20 = 0.6.  

For 6-10, RF = 3/20 = 0.15

For 11-15, RF = 2/20 = 0.1

For 16-20, RF = 1/20 = 0.05

For 21-25, RF=1/20 = 0.05

For 26-30, RF = 0/20 = 0.00

For 31-34, RF= 1/20 = 0.05.

To compute the cumulative frequency, add the existing frequency of each class with the previous frequencies.

For 1-5, CF = 0+12 = 12

For 6-10, CF = 12+3=15

For 11-15, CF = 12+3+2 = 17

For 16-20, CF=12+3+2+1 = 18

For 21-25, CF = 12+3+2+1+1=19

For 25-30, CF = 12+3+2+1+1+0=19

For 31-34, CF = 12+3+2+1+1+0+1=20

Now, to compute relative cumulative frequency, add the existing relative frequency of each class with the previous relative frequencies.  

For 1-5, RCF = 0+0.6

For 6-10, RCF = 0.6 + 0.15 = 0.75

For 11-15, RCF = 0.6+0.15+0.1 = 0.85

The rest of the relative cumulative frequencies can be computed in the same way.

Class(Minutes)           F       RF     CF     RCF

      1-5                       12     0.60    12     0.60  

      6-10                      3     0.15     15     0.75  

      11-15                      2     0.10     17     0.85  

      16-20                    1      0.05    18     0.90  

      21-25                    1      0.05    19     0.95  

      26-30                   0     0.00    19     0.95  

      31-34                     1     0.05    20       1  

       <u>Total                  20</u>  

Part 2. Now, we are asked to compute the Ogive graph which is also called as the cumulative frequency graph. The cumulative frequency needs to be plotted on the y-axis and the upper limit of each class needs to be plotted on the x-axis. The graph is attached.

(a) From the graph we can see that the number of workers who spend less than 5 minutes on unsolicited e-mail and spam are 12. So the answer for this part is 12 workers.

Percentage = 12/20 x 100 = 60%

(b) From the graph we can see that the number of workers who spend less than 10 minutes on spam e-mail are 15. The question is asking for the number of people who spend more than 10 minutes. For this we need to subtract 15 from the total number of workers.  

Number of workers spending more than 10 minutes = 20-15 = 5 workers.

Percentage = 5/20 x 100 = 25%

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3 years ago
PLEASE ANSWER ASAP!!
SCORPION-xisa [38]

Answer:Graph A

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3 years ago
If the subjective probability that it will rain is 45%, what is the probability that it will not rain
vovikov84 [41]

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There is a 55% chance that it won't rain.

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3 years ago
Use the recursive formula to answer the question.
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Step-by-step explanation:

7 0
3 years ago
the perimeter of a semicircle is 108 cm what is the perimeter of the square drawn on the diameter of the semicircle ​
Andreas93 [3]

Answer:

Step-by-step explanation:

the perimeter of the semi-circle would be the diameter plus the circumference of half of the circle.  

They want to know the perimeter of a square using the diameter of the sem-circle as ONE side, so the perimeter of the square would be 4 times the ONE side.

We should recall:

diameter = 2 times the radius     circumference of a cirlce = 2π r

How do we find the diameter of the of the semi circle?

The perimeter of the semi circle is given as 108 cm

Perimeter of the semicirle =    2r + π r         diameter plus semi circumference

                                  108   =   r ( 2 + π)           factor out the r and solve for r

                     108 / (2 + π)   =   r                       divide both sides by ( 2 + r)

Now we know r, the perimeter of the squqre is 4 times 2r  or  8r

           perimeter of square = 8  [ 108 / (2 + π) ]               π  I used 3.14

                                             =  864 / 5.14

                                             =   168.1 cm       I got rounded to nearest tenth    

<em>When I re-checked by work I found a few math,  logic and calculation errors.  Please re-check my answer for any more mistakes.</em>

       

4 0
3 years ago
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