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gtnhenbr [62]
3 years ago
13

A store is instructed by corporate headquarters to put a markup of ​% on all items. An item costing ​$ is displayed by the store

manager at a selling price of ​$. As an​ employee, you notice that this selling price is incorrect. Find the correct selling price. What was the​ manager's likely​ error? The correct selling price is ​$ 10.01 10.01. ​(Round to the nearest dollar as​ needed.)
Mathematics
1 answer:
natita [175]3 years ago
8 0

Answer:

hindi ko alam hehe

Step-by-step explanation:

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(3x+16) ( 5x-54) find the value of the x
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Answer:

15x^2 -82x-864

Step-by-step explanation:

MUse foil method

5 0
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)

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

#SPJ1

3 0
1 year ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%7B72x%5E%7B7%7D%20q%5E%7B8%7D%20%7D" id="TexFormula1" title="\sqrt{72x^{7} q^{8} }" al
Jet001 [13]
Attachment picture is the solution. Please mark me brain if you think it help. Thank you!

8 0
2 years ago
Please may you do the whole page for 30 points thanks
Ostrovityanka [42]
4 A) 1/6
4 B) 1/45
4 C) 2/3
4 D) 3/5
5 0
3 years ago
How do you type 40:16 ratio in to simplest form
Blizzard [7]
5:2
because you need to divide them by 8 to get to simplest form
7 0
3 years ago
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