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3 years ago

5

Define a variable and write an inequality to model each situation. A light bulb can be no more than 75 watts to be safely used i

n this light fixture.

1 answer:

DaniilM [7]3 years ago

3 0

Let the number of watts =w ‘no more than’ means less than or equal to, thus we will use that sign. The inequality can be written as: w<75

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Accuracy in taking orders at a drive-through window is important for fast-food chains. Periodically, QSR Magazine publishes "The

pav-90 [236]

Answer:

a) 0.7412 = 74.12% probability that all the three orders will be filled correctly.

b) 0.0009 = 0.09% probability that none of the three will be filled correctly

c) 0.0245 = 2.45% probability that at least one of the three will be filled correctly.

d) 0.9991 = 99.91% probability that at least one of the three will be filled correctly

e) 0.0082 = 0.82% probability that only your order will be filled correctly

Step-by-step explanation:

For each order, there are only two possible outcomes. Either it is filled correctly, or it is not. Orders are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The percentage of orders filled correctly at Burger King was approximately 90.5%.

This means that $p = 0.905$

You and 2 friends:

So 3 people in total, which means that $n = 3$

a. What is the probability that all the three orders will be filled correctly?

This is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{3,3}.(0.905)^{3}.(0.095)^{0} = 0.7412$

0.7412 = 74.12% probability that all the three orders will be filled correctly.

b. What is the probability that none of the three will be filled correctly?

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.905)^{0}.(0.095)^{3} = 0.0009$

0.0009 = 0.09% probability that none of the three will be filled correctly.

c. What is the probability that one of the three will be filled correctly?

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{3,1}.(0.905)^{1}.(0.095)^{2} = 0.0245$

0.0245 = 2.45% probability that at least one of the three will be filled correctly.

d. What is the probability that at least one of the three will be filled correctly?

This is

$P(X \geq 1) = 1 - P(X = 0)$

With what we found in b:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0009 = 0.9991$

0.9991 = 99.91% probability that at least one of the three will be filled correctly.

e. What is the probability that only your order will be filled correctly?

Yours correctly with 90.5% probability, the other 2 wrong, each with 9.5% probability. So

$p = 0.905*0.095*0.095 = 0.0082$

0.0082 = 0.82% probability that only your order will be filled correctly

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Step-by-step explanation:

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45-38= 7min decrease
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