. 5.8x10 to the power of 7
1 answer:
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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.
a)
A number of extension lines needed to accommodate $90 in calls immediately:
Use the calculation for busy k servers.
The probability that 2 servers are busy:
The likelihood that 2 servers will be busy may be calculated using the formula below.
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.
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.
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,
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.
For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.
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Answer:
(1.5, -13.5)
Step-by-step explanation:
Midpoint Formula:
Simply plug in our coordinates into the formula:
x = (17 - 14)/2
x = 3/2
y = (-11 - 16)/2
y = -27/2