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2 years ago

500 portions of strawberries have been ordered for the tournament, but because of bad weather only

Mathematics

1 answer:

Citrus2011 [14]2 years ago

8 0

We conclude that 375 portions can be sold.

<h3>How many portions can be sold?</h3>

We know that 500 portions have been ordered, but only 5/6 of that was delivered, so the number of delivered portions is:

p = (5/6)*500 = 416.6

Which we can round to the next whole number, 417.

Now we know that 10% of these are given to the players and staff, then the number of portions that can be sold is the 90% of 417, which is:

N = 417*(90%/100%) = 417*0.9 = 375

We conclude that 375 portions can be sold.

If you want to learn more about percentages:

brainly.com/question/843074

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3 years ago

The probability that a certain make of car will need repairs in the first seven months is 0.9. A dealer sells three such cars. W

aleksandr82 [10.1K]

Answer:

0.9990 = 99.90% probability that at least one of them will require repairs in the first seven months.

Step-by-step explanation:

For each car, there are only two possible outcomes. Either they will require repair in the first seven months, or they will not. The probability of a car requiring repair in the first seven months is independent of other cars. So 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 probability that a certain make of car will need repairs in the first seven months is 0.9.

This means that $p = 0.9$ A dealer sells three such cars.

A dealer sells three such cars.

This means that $n = 3$

What is the probability that at least one of them will require repairs in the first seven months?

Either none will require repairs, or at least one will. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X > 0) = 1$

We want P(X > 0). So

$P(X > 0) = 1 - P(X = 0)$

In which

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

$P(X = 0) = C_{3,0}.(0.9)^{0}.(0.1)^{3} = 0.001$

Then

$P(X > 0) = 1 - P(X = 0) = 1 - 0.001 = 0.9990$

0.9990 = 99.90% probability that at least one of them will require repairs in the first seven months.

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3 years ago