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3 years ago

Write (-3, -5) (-1, -19) in point slope form

Mathematics

1 answer:

stiv31 [10]3 years ago

8 0

Answer:

(y + 5) = -7(x + 3)

Step-by-step explanation:

Point Slope Form: (y - y1) = m(x - x1)

<u>Step 1: Find Slope</u>

m = $\frac{y2 - y1}{x2 - x1}$

m = $\frac{-19 - (-5)}{-1 - (-3)}$

m = $\frac{-19 + 5}{-1 + 3}$

m = $\frac{-14}{2}$

m = -7

<u>Step 2: Put into point slope form</u>

Point Slope Form: (y - y1) = m(x - x1)

(y - (-5)) = -7(x - (-3))

(y + 5) = -7(x + 3)

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The average low temperature by month in Nashville is represented by the function

malfutka [58]

Answer:

Step-by-step explanation:

Given the average low temperature by month in Nashville is represented by the function f(x)=-1.4x² + 19x +1.7, where x is the month, the average rate of change is expressed as d[f(x)]/dx = 2(1.4x) + 19

d[f(x)]/dx = 2.8x + 19

Since the number of months between March and August is 5 months and x is in months, hence we will substitute x = 5 into the resulting function to get the average rate of change from March to August as shown;

d[f(x)]/dx at x = 5

= 2.8(5)+ 19

= 14 + 19

= 33

<em>Hence the average rate of change from March to August is 33</em>

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3 years ago

The area of a square is 20 square which best represent the length of the side of the square

mote1985 [20]

5' per side or whatever units it's measured in, because 4x5=20

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( p^2 - 8p^4 - 11p^3 ) + ( 4p^2 + 7p^4 )

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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 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.

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

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.