What is true about the interior and exterior angle measures of any triangle

A(0, 1), B(4,1), C(1, 2), D(1,6): AB and CD

Ivanshal [37]

The answer is D because I took the test

8 0

3 years ago

The ages of trees in a forest are normally distributed with a mean of 25 years and a standard deviation of 4. Approximately what

harkovskaia [24]

Answer:

78.88%

Step-by-step explanation:

We have been given that

$\mu=25,\sigma=4,x_1=20,x_2=30$

The z-score formula is given by

$z-\text{score}=\frac{x-\mu}{\sigma}$

For $x_1=20$

$z_1=\frac{20-25}{4}\\\\z_1=-1.25$

For $x_2=30$

$z_2=\frac{30-25}{4}\\\\z_2=1.25$

Now, we find the corresponding probability from the standard z score table.

For the z score -1.25, we have the probability 0.1056

For the z score 1.25, we have the probability 0.8944

Therefore, the percent of the trees that are between 20 and 30 years old is given by

0.8944 - 0.1056

= 0.7888

=78.88%

6 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%