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ASHA 777 [7]
3 years ago
8

Solve the problems.

Mathematics
1 answer:
RoseWind [281]3 years ago
6 0

Answer:

????

Step-by-step explanation:

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A marching band stands in a triangle formation. The drum major stands at A, one of the vertices of the triangle. The whole forma
iragen [17]

Answer:

D. (1, 3)

Step-by-step explanation:

Based on the image uploaded, during the first march, move point A, 3 units left, the new position will (3, -4), then move point A again 5 units up, the new position will be (3,1).

From second march move point A  2 units left, the new position will be (1,1), finally move point A 2units up, the new position and last position will be (1, 3).

Thus, the drum major be standing on (1, 3)

7 0
3 years ago
Mid-West Publishing Company publishes college textbooks. The company operates an 800 telephone number whereby potential adopters
s344n2d4d5 [400]

The various answers to the question are:

  • To answer 90% of calls instantly, the organization needs four extension lines.
  • The average number of extension lines that will be busy is Four
  • For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

<h3>How many extension lines should be used if the company wants to handle 90% of the calls immediately?</h3>

a)

A number of extension lines needed to accommodate $90 in calls immediately:

Use the calculation for busy k servers.

$$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$$

The probability that 2 servers are busy:

The likelihood that 2 servers will be busy may be calculated using the formula below.

P_{2}=\frac{\frac{\left(\frac{20}{12}\right)^{2}}{2 !}}{\sum_{i=0}^{2} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$

Hence, two lines are insufficient.

The probability that 3 servers are busy:

Assuming 3 lines, the likelihood that 3 servers are busy may be calculated using the formula below.

P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{2} \frac{\left(\frac{\lambda}{\mu}\right)^{i}}{i !}}$ \\\\$P_{3}=\frac{\frac{\left(\frac{20}{12}\right)^{3}}{3 !}}{\sum_{i=0}^{3} \frac{\left(\frac{20}{12}\right)^{1}}{i !}}$$\approx 0.1598$

Thus, three lines are insufficient.

The probability that 4 servers are busy:

Assuming 4 lines, the likelihood that 4 of 4 servers are busy may be calculated using the formula below.

P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$ \\\\$P_{4}=\frac{\frac{\left(\frac{20}{12}\right)^{4}}{4 !}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{7}}{i !}}$

Generally, the equation for is  mathematically given as

To answer 90% of calls instantly, the organization needs four extension lines.

b)

The probability that a call will receive a busy signal if four extensions lines are used is,

P_{4}=\frac{\left(\frac{20}{12}\right)^{4}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{1}}{i !}} $\approx 0.0624$

Therefore, the average number of extension lines that will be busy is Four

c)

In conclusion, the Percentage of busy calls for a phone system with two extensions:

The likelihood that 2 servers will be busy may be calculated using the formula below.

P_{j}=\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}$$\\\\$P_{2}=\frac{\left(\frac{20}{12}\right)^{2}}{\sum_{i=0}^{2 !} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$

For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

Read more about signal

brainly.com/question/14699772

#SPJ1

3 0
2 years ago
Anybody Know This. ?
FinnZ [79.3K]

Answer:

No... i'm sorry

Step-by-step explanation:

4 0
2 years ago
What is the quotient of the following in simplest form? 5/9 divided by 2/3
Assoli18 [71]

Answer:

5/6

Step-by-step explanation:

You write the question as such: 5/9 divided by 2/3.

After that, you keep the first fraction, flip the sign and the fraction.

You will get: 5/9 multiplied by 3/2.

When you get this, you then cross multiply and simplify as such.

5/3 multiplied by 1/2.

5/6 would be your solution.

4 0
2 years ago
Can someone please help me on this
Naddik [55]
12345678911111111111
7 0
3 years ago
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