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Doss [256]
3 years ago
15

A shipment of 12 microwave ovens contains 4 defective units. A restaurant buys three of these units. What is the probability of

the restaurant buying at least two nondefective​ units?
Mathematics
1 answer:
hammer [34]3 years ago
5 0
The <u>correct answer</u> is:

20/27 = 0.7407.

Explanation:

To find the probability that the restaurant buys at least 2 non-defective units, we find the probability that they purchase either 2 or 3 non-defective units:

P(r) = _nC_r(p)^r(1-p)^{n-r}&#10;

They purchase <u>3 units total</u>, so this is <u>n</u>.
We want the <u>probability of either 2 or 3</u> non-defective units; this means <u>r is 2 and r is 3</u>.
In this case we want the probability that the units are non-defective.  Since there are 4 defective units, this means there are 12-4=8 non-defective units; this makes the <u>probability of a non-defective unit 8/12 = 2/3.  This is p</u>.

Using these, we have:
P(X=2 or X=3)=_3C_2(\frac{2}{3})^2(1-\frac{2}{3})^{3-2}+_3C_3(\frac{2}{3})^3(1-\frac{2}{3})^{3-3}&#10;\\&#10;\\=\frac{3!}{2!1!}(\frac{2}{3})^2(\frac{1}{3})^1+\frac{3!}{3!0!}(\frac{2}{3})^3(\frac{2}{3})^0&#10;\\&#10;\\=3(\frac{4}{9})(\frac{1}{3})+1(\frac{8}{27})(1)&#10;\\&#10;\\=\frac{12}{27}+\frac{8}{27}=\frac{20}{27}=0.7407
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THIS ANSWER IS FOR f(3):
f(x)=x+3
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THIS ANSWER IS FOR f(-2):
f(x)=x+3
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7 0
2 years ago
What is the radius, in centimeters, of a circle that has a circumference of 161 centimeters?
elena-14-01-66 [18.8K]

Answer:

25.63694268 = r

Step-by-step explanation:

The circumference is

C = 2*pi*r

161 = 2*(3.14) *r

161 = 6.28 r

Divide each side by 6.28

161/6.28 = r

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3 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

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3 0
1 year ago
I need help with this question!!!!
Genrish500 [490]

Answer: The answer is A over B if you know what im sayin

Step-by-step explanation:

You take A and B and put the A over the B.

7 0
2 years ago
Which relations are functions?<br><br><br><br> Choose all answers that are correct.
-Dominant- [34]
Hello!

In a function, each input has only one output. In A, three has two outputs, 4 and 5, so A is not a function.

In B, you can use something called the vertical line test to see if each x value has one y value as an output. You move an imaginary vertical line across the graph, and if it intersects with two points it is not a function. If we do this on our graph, it will not intersect two points. Therefore, B is a function.

In C, we can see that each input has one output, or there are all different inputs, so C is a function.

For D we can use that vertical line test again. It intersects both the points (-1,1) and (-1,6) so D is not a function

Our final answers are B and C.


I hope this helps!




6 0
3 years ago
Read 2 more answers
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