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3 years ago

13

Mr Ruiz drove 305 miles in 5 hours on Saturday and 190 miles in 4 hours on Sunday what is his average speed in miles per hour fo

r the two days

1 answer:

Snezhnost [94]3 years ago

4 0

Add the mileage together to find total miles.

Add the hours together.

Divide total miles by total hours.

305 + 190 = 495 miles

5 + 4 = 9 hours.

495 / 9 = 55 miles per hour.

The average speed was 55 miles per hour.

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

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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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Answer:

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If 5 friends are sharing the berries, how many pounds of berries does each friend receive? Is the answer to 3/4 divided by 2/5 greater than or less than 1.

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