All categories

3 years ago

NEED HELP ASAP PLEASE HELP ASAP THANKS 15 Points

Mathematics

1 answer:

11111nata11111 [884]3 years ago

4 0

<h3>Answer: 224</h3>

To get this answer, you simply multiply out all the values given. So we have 2*4*7*4 = 8*28 = 224. The reason why this works is to imagine we only had two aspects to focus on. Let's say we could only choose 2 sizes and 4 colors. Lay out a table that has 2 rows and 4 columns. This table will have 2*4 = 8 different cells inside it to represent the 8 different combinations possible. This idea can be extended to more dimensions than just two, though it might be tricky to visualize. Another way to see this is to draw out a tree diagram. Unfortunately with large numbers such as this, the tree diagram gets very unwieldy. In your math book, this concept may be referred to as the "counting principle".

You might be interested in

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.

Read more about signal

brainly.com/question/14699772

#SPJ1

3 0

2 years ago

Expand 4(x+3)+6(x+2)

ikadub [295]

Expanded=4x+12+6x+12 Simplified=10x+24

3 0

3 years ago

Read 2 more answers

Talisa plans a 30-foot deep pond. While digging, she hits rock 25 feet down. How can Talisa modify

Natali [406]

9514 1404 393

Answer:

multiply the radius by √1.2 ≈ 1.0954

Step-by-step explanation:

Let the original depth and radius be represented by 30 and r. Let the modified depth and radius be represented by 25 and r'. Assuming the pond is uniform depth, the formula for the volume of a cylinder applies. The volume of a cylinder is given by ...

V = πr²h

In this application, we want the original and modified volumes to be the same.

πr²(30) = π(r')²(25)

(r')² = r²(30/25) = r²(6/5)

r' = r√(6/5)

The radius can be multiplied by √1.2 to maintain the original volume of the pond.

6 0

3 years ago

Solution A is 30% alcohol and solution B is 60% alcohol. How much of each is needed to make 80 liters of a solution that is 45%

postnew [5]

We need 80 liters of 45% alcohol
We have 30% alcohol and 60% alcohol.
This is relatively easy. We need equal amounts of each.
There fore we mix 40 liters of 30% alcohol and 40 liters of 60% alcohol.

7 0

4 years ago

Evaluate 6C3. <br><br> 18<br> 20<br> 60<br> 120

melomori [17]

Answer:

Step-by-step explanation:

$nC_{r}=\frac{n!}{r!(n-r)!}\\\\6C_{3}=\frac{6!}{3!(6-3)!}\\\\=\frac{6!}{3!(6-3)!}==\frac{6!}{3!3!}\\=\frac{6*5*4*3!}{3!3!}\\\\=\frac{6*5*4}{3!}\\\\=\frac{6*5*4}{3*2*1}\\\\ =5*4=20$

4 0

3 years ago