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NARA [144]
2 years ago
14

Find the sum of -8x^2+3 and -9x^2+10x-10

Mathematics
1 answer:
nirvana33 [79]2 years ago
3 0

Answer:

-17x² + 10x - 7

General Formulas and Concepts:

<u>Algebra I</u>

Terms/Coefficients

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

-8x² + 3 - 9x² + 10x - 10

<u>Step 2: Simplify</u>

  1. Combine like terms (x²):                                                                                   -17x² + 3 + 10x - 10
  2. Combine like terms:                                                                                         -17x² + 10x - 7
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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

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3 0
1 year ago
Determine the value for x in the given diagram when a||b.
Sladkaya [172]

Answer:

55°

Step-by-step explanation:

Since a||b

Then corresponding angles are equal

So 55 is correspondent to X

Therefore X=55°

4 0
3 years ago
Read 2 more answers
30 feet below the surface of the water. You are ascending towards the
fomenos

Answer is slope of 5


Step by step


You can graph the slope as seen on the attached picture. You go up 30, you go right +6. Slope is y/x, so 30/6, equals 5



You can also use the slope formula


(y2-y1) over (x2-x1)


Using points on the graph


(6,0) and (0,-30)


(-30-0 ) over (0-6)


-30 over -6


-30/-6


= 5


5 0
2 years ago
PLEASE help with these 2 questions<br><br> thank you!
Sergio039 [100]
Download the app photo math help a whole lot
6 0
3 years ago
what is the relationship between 5.34 x 10^5 and 5.34 x 10^-2 is it greater or lesser than? for 5.34 x 10^5? will give brainlies
Romashka-Z-Leto [24]

the relationship between 5.34 x 10^5 and  5.34 x 10^-2

5.34 x 10^5 = 5.34*100000= 534000

5.34 * 10^-2  , for exponent -2 we move the decimal point 2 places to the left

5.34 x 10^-2= 0.0534

Now we compare 534000  and 0.0534

0.0534 is 10,000,000 times less than 534000

Or we can say  534000 is 10,000,000 times greater than 0.0534

5.34 x 10^5 is 10,000,000 times greater than 5.34 x 10^-2


3 0
3 years ago
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