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2 years ago

Giving 97 points for answer. Need asap for a, b, and c pls help.

Mathematics

2 answers:

Kay [80]2 years ago

4 0

Answer:

sabi 97 tas 49 lang amp ginagawa mo

SIZIF [17.4K]2 years ago

3 0

Answer:

2. being vertically opposite angle.

3.being alternate angle

4.because these points never meet . and they are cut by a transversal line

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A survey found that women's heights are normally distributed with mean 62.8 inches and standard deviation of 2.5 inches. This su

Jet001 [13]

a)

Z*_Upper = (76 - 62.7)/2.5 = 5.32

Z*_Lower = (57 - 62.7)/2.5 = -2.28

The requirement is to get p(-2.28 < Z < 5.32) = p(Z<5.32) - p(Z<-2.28).

Use normal distribution table to get the answer for p and multiply with 100 to get the percentage.

The other questions are now easy for you to answer on your own. Hope it helps.

4 0

2 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

Total 20

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%