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3 years ago

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Faucet A flows at a rate of 100 liters per 5 minutes and Faucet B flows at a rate of 100 liters per 4 minutes into an empty pool

. How many minutes do the two faucets take to fill a pool with 9,000 liters of water?

1 answer:

Dmitriy789 [7]3 years ago

8 0

Answer:

It will take both faucets 200 minutes to fill a pool with 9,000 liters of water

Step-by-step explanation:

Here, we want to know the time it will take for both faucets to fill a pool with 9,000 liters of water

What we need here is to have a joint rate for both faucets;

For the first faucet, we have a rate of 100 liters per 5 minutes so this means in a minute;

100/5 = 20 liters per minute

For the second faucet, the per minute rate will be 100/4 = 25 liters per minute

So therefore, the joint rate will be;

25 liters per minute + 20 liters per minute = 45 liters per minute

So the time needed to fill 9,000 liters of water in a pool will be ;

9000 liters/ 45 liters per minute = 200 minutes

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1 year ago

A food-packaging apparatus underfills 10% of the containers. Find the probability that for any particular 10 containers the numb

Maksim231197 [3]

Answer:

a) $P(X = 1) = 0.38742$

b) $P(X = 3) = 0.05740$

c) $P(X = 9) = 0.00000$

d) $P(X \geq 5) = 0.00163$

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it is undefilled, or it is not. This means that we can solve this problem using the binomial probability distribution.

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 combinatios 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.

In this problem

There are 10 containers, so $n = 10$ .

A food-packaging apparatus underfills 10% of the containers, so $p = 0.1$ .

a) This is P(X = 1)

$P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742$

b) This is P(X = 3)

$P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740$

c) This is P(X = 9)

$P(X = 9) = C_{10,9}.(0.1)^{9}.(0.9)^{1} = 0.00000$

d) This is $P(X \geq 5)$ .

Either the number is lesser than five, or it is five or larger. The sum of the probabilities of each event is decimal 1. So:

$P(X < 5) + P(X \geq 5) = 1$

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

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

$P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.34868$

$P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742$

$P(X = 2) = C_{10,2}.(0.1)^{2}.(0.9)^{8} = 0.1937$

$P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740$

$P(X = 4) = C_{10,4}.(0.1)^{1}.(0.9)^{9} = 0.38742$

So

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.34868 + 0.38742 + 0.19371 + 0.05740 + 0.01116 = 0.99837$

Finally

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.99837 = 0.00163$

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