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pshichka [43]
2 years ago
9

Hello help me please​

Mathematics
2 answers:
laila [671]2 years ago
5 0

Answer:

7

Step-by-step explanation: it intersects on the y-axis

Kaylis [27]2 years ago
5 0
The answer is 7 because
It is along the y-axis and 0 is the x intercept and 7 is y intercept.
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Mid-West Publishing Company publishes college textbooks. The company operates an 800 telephone number whereby potential adopters
Mumz [18]

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 extension lines

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8 0
1 year ago
Evaluate the expression 12 - x when x = 4. Show your work.
ArbitrLikvidat [17]

Answer: 8

Step-by-step explanation: 12 minus 4 equals 8

6 0
2 years ago
Read 2 more answers
Using b=1+r, what does b=
alisha [4.7K]

Answer:

Adding 12 to the circle area is equal to the square area.

Or

s2 = 12 + A

Where

s = side of square

A = area of circle

So

s2 = 12 + 36

s2 = 48

Solve this for s to get the side length

4 0
2 years ago
A student writes an incorrect step while checking if the sum of the measures of the two remote interior angles of triangle ABC b
blsea [12.9K]
The correct answer is A.  All triangles' interior angles equal 180.
5 0
3 years ago
Read 2 more answers
Use the graph to determine
Sonja [21]

Answer:

a. Domain : All Real Numbers

b. Range: y≥ -1

c. The x-intercepts are (0,0) and (-4,0)

d. The y-intercept is (0,0).

e. f(2)=3

   f(-2)=-1

Step-by-step explanation:

If this has helped you please mark as brainliest

3 0
3 years ago
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