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stealth61 [152]
3 years ago
13

Nadia is a stockbroker. She earns 13​% commission each week. Last​ week, she sold ​$4,500 worth of stocks. How much did she make

last week in​ commission? If she averages that same amount each​ week, how much did she make in commission in​ 2011?
Mathematics
1 answer:
m_a_m_a [10]3 years ago
6 0

Multiply amount sold by commission rate:

4500 x 0.13 = 585

She made $585 last week.

Multiply amount made in 1 week by 52 weeks ( 1 year = 52 weeks)

585 x 52 = 30,420

She made $30,420 in 2011

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Mid-West Publishing Company publishes college textbooks. The company operates an 800 telephone number whereby potential adopters
Mumz [18]

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 extension lines

brainly.com/question/13362603

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8 0
2 years ago
What is the answer??
Amanda [17]

Answer:

angle 1= 104                angle 2= 76

Step-by-step explanation:

angle 1 equals x+66 and angle 2 equals 2x so you set up an equation where they are the sum of 180 degrees because the angles are supplementry angles.

x+66+2x=180                        set the sum of the angles to 180

3x+66=180                            combind like terms

3x=144                                   subract 66 from both sides

x=38                                      divide 3 from both sides

Now you put x back into the measurments

angle 1= (x+66)                              angle 2= 2x

              (38+66)                                           2×38

angle 1= 104                                   angle 2= 76

Or after you know angle 1 you can subtract angle 1 from 180 to get angle 2.

5 0
3 years ago
Does anybody like anime ??
ad-work [718]
Of course bro why wouldn’t we like anime
3 0
3 years ago
HELP!!!!!!!!! I DONT UNDERSTAND THIS
ValentinkaMS [17]

Answer:

x=86^{\circ}

y=94^{\circ}

z=54^{\circ}

Step-by-step explanation:

First, let's look at the largest triangle (two smaller triangles are combined) to solve for z. Since the sum of the angles in a triangle adds up to 180^{\circ}, we can write the equation:

20+74+32+z=180

z=180-126

z=54^{\circ}

Looking at the smaller triangle on the left, y (exterior angle) is the sum of the two opposite interior angles of the triangle on the left:

y=74+20

=94^{\circ}

Since x is the exterior angle of the triangle on the right, it is equivalent to the sum of the opposite interior angles of that triangle:

x=32+z

=32+54

=86^{\circ}

Hope this helps :)

6 0
3 years ago
Which choice shows the coordinates of C' if the trapezoid is reflected across the y-axis?
nirvana33 [79]

Answer:

(5,3)

Step-by-step explanation:

7 0
3 years ago
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