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Tatiana [17]
3 years ago
14

Q # 15 please somebody help me

Mathematics
2 answers:
djyliett [7]3 years ago
5 0

<u>Answer</u>

7m + 9

<u>Explanation</u>

1/3 (21m + 27)             First open the brackets by multiplying what is inside the parenthesis by 1/3.

1/3 (21m + 27)  = 1/3×21m + 1/3×27

                       = 7m + 9

Mumz [18]3 years ago
3 0

1/3(21m + 27) =

= 1/3 * 21m + 1/3 * 27

= 7m + 9

Answer: B. 7m + 9

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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
4 years ago
Look at the picture please help
morpeh [17]

Answer:

c

Step-by-step explanation:

First off, we know that (g ° f)(2) is just both functions done in a ordered fashion. In this case, f(x) is done first.

To figure out what f(2) is, all we have to do is find where x = 2 is on the graph. In this case it is on point (2, 3). The input is the x and the output is the y, so f(2) = 3.

Then, we can figure out what g(3) by locating x = 3 on line g. It shows the point (3, 0). This means that g(3) = 0.

3 0
3 years ago
Please answer these 2 questions please!
Leno4ka [110]

Answer:

The 4th answer, at the bottom.

Step-by-step explanation:

The first answer is proportional, because y is equivalent to .5x.

The second answer is also proportional, because the ratios are equivalent.

The third answer is also proportional, because the line goes through the origin.

The fourth answer isn't proportional, because the ratios aren't equivalent.

5 0
4 years ago
What is the solution set for this inequality?
Ostrovityanka [42]

Answer:

X<-4

Step-by-step explanation:

-3x+10>22

subtract 10 to both sides

-3x>12

divide by -3 on both sides

when you divide by a negative number you flip the sign so the answer will be x<-4

8 0
3 years ago
A 14 gram sample of a substance that's used to preserve fruit and vegetables has a k-value of 0.1092. Find the substance's half-
nignag [31]

Answer:

t = 6.3

Step-by-step explanation:

N=Noe^(-kt)

No = 14 grams

k = .1092

We want to find t when N = 7 or 1/2 of 14

N=Noe^(-kt)

7 = 14 e ^ (-.1092t)

Divide each side by 14

1/2 = e ^ (-.1092t)

take the natural log of each side

ln (1/2) = ln e ^ (-.1092t)

ln (1/2) = -.1092t

Divide each side by -.1092

ln (1/2)/ -.1092 = t

t≈6.3475

Rounding to the nearest tenth

t = 6.3

5 0
4 years ago
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