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2 years ago

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Emily’s last water bill listed a previous reading of 7,123 ccf and a present reading of 7,171 ccf. Her water company charges $0.

73 per ccf of water. What should Emily have been charged on her water bill?

1 answer:

ludmilkaskok [199]2 years ago

8 0

It is given in the question that

Emily’s last water bill listed a previous reading of 7,123 ccf and a present reading of 7,171 ccf. Her water company charges $0.73 per ccf of water.

Therefore number of ccf is

$= 7171-7123 = 48ccf$

And company charges $0.73 per ccf.

So for 48 ccf, company charges

$= 0.73*48 = $35.04$

And that's the required answer .

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Answer:

0.4 = 40% probability that both marbles are blue.

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Step-by-step explanation:

To solve this question, we need to understand conditional probability(for the final question) and the hypergeometric distribution(for the first three, because the balls are chosen without being replaced).

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N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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

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

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We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

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4 + 2 = 6 total marbles, which means that $N = 6$

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Sample of 2, which means that $n = 2$

This is P(X = 2). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

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What is the probability that both marbles are yellow?

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