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NNADVOKAT [17]
3 years ago
15

Write the number equal to 5 tens and 13 ones.

Mathematics
1 answer:
tekilochka [14]3 years ago
3 0
Hi, Sthefy! This is just a little bit of adding, not too hard! Okay, 5 tens. What is 1 ten equal to? 10!
 
2 tens? 20!

3 tens? 30!
 
You get the idea? Okay, so now we know that 5 tens are 50. Now, ones. All we need to do now is add 13 + 50! Let's break it down, make it easier.

13.. Hm, what can we do to make this easier? 

Ah! Take away the 3 from 13 for now. So just 10 + 50. That's 60. Now, add the 3. 63! That wasn't hard, was it? I hope I helped you!
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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%

4 0
3 years ago
What type of probability uses a knowledge of sample spaces as opposed to observations to determine the numerical probability of
Aneli [31]

Answer:

Classical probability

Step-by-step explanation:

  • Classical property is that type of probability or statistical concept that deals with the  measurement of the odds or likelihood of some event.
  • It is that form of probability that is simplest in approach and includes the equal likelihood of happening or occurrence of an event.
  • In order to determine the numerical probability of an event, it makes use of the knowledge of sample spaces in contrast with the observations.
  • It takes a proper note of an event in order to calculate its probability and also it deals with the likelihood of a certain event rather than making observations.

5 0
3 years ago
Read 2 more answers
What is 0.32 of 44???​
vladimir1956 [14]
14.08 that’s if in this case “of” means multiply hope this helps
8 0
2 years ago
Read 2 more answers
Did I find the area of the composite figures correctly in the pic? If not, help pls?
Annette [7]
No, in problem 1
you said that area of triangle=bh when it should be 1/2bh
the area should be 9 not 18 so triange=9
aera of rectangle is correct
so
27+9=36 in^2 is answer



2. it is 1 rectangle 1 triangle 1 square  you forgot the square
rectangle=12 times 4=48
square=4 times 4=16
triangle=1/2bh=1/2(8)4=16
add

48+16+16=80 in^2


remember that
triangle=1/2(base times height)
draw immaginary lines and split shapes, it could help to seperate shapes or/and mmove them around and regroup them
8 0
3 years ago
Please help me asap<br>​
Lera25 [3.4K]

Answer:

Step-by-step explanation:

From the picture attached,

∠4 = 45°, ∠5 = 135° and ∠10 = ∠11

Part A

∠1 = ∠4 = 45°  [Vertically opposite angles]

∠1 + ∠3 = 180° [Linear pair of angles]

∠3 = 180° - ∠1

     = 180° - 45°

     = 135°

∠2 = ∠3 = 135°  [Vertically opposite angles]

∠8 = ∠5 = 135° [Vertically opposite angles]

∠5 + ∠6 = 180° [Linear pair of angles]

∠6 = 180° - 135°

∠6 = 45°

∠7 = ∠6 = 45° [Vertically opposite angles]

By triangle sum theorem,

m∠4 + m∠7 + m∠10 = 180°

45° + 45° + m∠10 = 180°

m∠10 = 180° - 90°

m∠10 = 90°

m∠10 = m∠12 = 90°  [Vertically opposite angles]

m∠10 = m∠11 = 90° [Given]

Part B

1). ∠1 ≅ ∠4  [Vertically opposite angles]

2). ∠7 + ∠5 = 180° [Linear pair]

3). ∠9 + ∠10 = 180° [Linear pair]

7 0
3 years ago
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