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Naily [24]
3 years ago
12

100 points, put nonsense, get reported. Susan wants to build a floor to put at the bottom of her tree house. She made the scale

drawing below using a scale of 2.5 in = 3 ft. Enter the length Susan must use for the floor.

Mathematics
2 answers:
Elanso [62]3 years ago
3 0

Divide the length of the scale drawing by the scale (2.5) then multiply that by the scale feet (3)


11 inches / 2.5 = 4.4

4.4 x 3 = 13.2 feet


Answer: 13.2 feet

AveGali [126]3 years ago
3 0
  • Scale is 2.5in

Total 2.5in pairs

\\ \tt\hookrightarrow \dfrac{11}{2.5}

\\ \tt\hookrightarrow 4.4

  • Now it is equal to 3ft

So length=4.4(3)=13.2ft

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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.

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2 years ago
2. Select the properties that both rectangles and parallelograms always share.
tatyana61 [14]

Answer:

Here you go! :)

2. 2 pairs of parallel sides

3. 3 sides of equal length

4 0
3 years ago
The perimeter of a rectangle is 72 m. The width of the rectangle is 16 m. What is the area of the rectangle
Aneli [31]

Answer:

area=320m²

Step-by-step explanation:

perimeter=2(l+w)

72=2(l+16)

72/2=l+16

36=l+16

36-16=l

l=20m

area=l*w

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 =320m²

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If the spinner is spun 40 times, predict the number of times the spinner would land on Section B
tester [92]

Answer:

11

Step-by-step explanation:

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3 0
4 years ago
write an equation in point-slope form for the line through the given point with the given slope. (10, –9); m = –2 y – 10 = –2(x
fomenos
Y - y₁ = m(x -x₁)

slope m = -2,  from point (10, -9) ,  x₁ = 10, y₁ = -9

 y - y₁ = m(x -x<span>₁)

</span>y - -9 = -2(x - 10<span>)</span><span>

y + 9 = -2(x - 10) </span>
7 0
3 years ago
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