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Kisachek [45]
2 years ago
13

Multiply the quotient of b and 10 by 9

Mathematics
1 answer:
Lisa [10]2 years ago
8 0

Answer:

10b/9

Step-by-step explanation:

multiply the quotient of b and 10 by 9

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To practice for a competition, Sonya swam 0.94 kilometer in the pool each day for 3 weeks. How many meters did Sonya swim in tho
Strike441 [17]

Answer:

Distance swam by Sonya in 3 weeks = 19740 m

Step-by-step explanation:

To practice for a competition, Sonya swam 0.94 kilometer in the pool each day for 3 weeks.

Number of days in 1 week = 7

Number of weeks she swam = 3

Total number of days she swam = 3 x 7 = 21 days

Distance swam per day = 0.94 km

Total distance swam = Total number of days she swam x Distance swam per day

Total distance swam = 21 x 0.94 = 19.74 km = 19.74 x 1000 m = 19740 m

Distance swam by Sonya in 3 weeks = 19740 m

7 0
3 years ago
If I could get some help it would be very much appreciated
raketka [301]

Answer:

Step-by-step explanation:

A circle is 360 degrees. What we are asked in words is, "How many degrees is 40% of 360?" Let x be "how many degrees"; the word "is" means =; 40% as a decimal is .4; the word of in this context means to multiply; and 360 is just 360. Putting all that together into an equation we can solve for x:

x = .4(360) so

x = 144

Adventure movies covers 144 degrees of that circle.

5 0
3 years ago
Read 2 more answers
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
Pls help 20 points if u answer
yarga [219]
I have absolutely no idea i hate math so much
3 0
2 years ago
Show if √5×√5×√8 is rational or irrational numbers​
kobusy [5.1K]
√5 x√5 x√8= √(5x5x8) = √200 = 10 √2
It’s an irrational number
8 0
1 year ago
Read 2 more answers
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