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mezya [45]
3 years ago
13

in the second half of the game, the Wildcats again scored 18 points. This time, however, we don't know how many total shots were

made. List all of the possible combinations of 2- and 3- point shots that could total 18 points. include a sentence explain how you know that you found all of the possible answers. please help me answer this would really appreciate it.
Mathematics
1 answer:
Kobotan [32]3 years ago
3 0

Lets start with the chain of 3's:

18 = 3 + 3 + 3 + 3 + 3 + 3          

But, we know that

3 + 3 = 2 + 2 + 2

Sol, let's replace 3 + 3 by 2 + 2 + 2 one by one.

Hence, the possible ways of combinations are listed below:

18 = 3 + 3 + 3 + 3 + 2 + 2 + 2                        (1)

18 = 3 + 3 + 2 + 2 + 2 + 2 + 2 + 2                  (2)

Therefore, there are two combinations of 2- and 3- point shots that could total 18 points.

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2 years ago
A popular resort hotel has 300 rooms and is usually fully booked. About 7% of the time a reservation is canceled before the 6:00
kicyunya [14]

Answer:

8.69% probability that at least 285 rooms will be occupied.

Step-by-step explanation:

For each booked hotel room, there are only two possible outcomes. Either there is a cancelation, or there is not. So we use concepts of the binomial probability distribution to solve this question.

However, we are working with a big sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

A popular resort hotel has 300 rooms and is usually fully booked. This means that n = 300

About 7% of the time a reservation is canceled before the 6:00 p.m. deadline with no pen-alty. What is the probability that at least 285 rooms will be occupied?

Here a success is a reservation not being canceled. There is a 7% probability that a reservation is canceled, and a 100 - 7 = 93% probability that a reservation is not canceled, that is, a room is occupied.  So we use p = 0.93

Approximating the binomial to the normal.

E(X) = np = 300*0.93 = 279

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.93*0.07} = 4.42

The probability that at least 285 rooms will be occupied is 1 subtracted by the pvalue of Z when X = 285. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{285- 279}{4.42}

Z = 1.36

Z = 1.36 has a pvalue of 0.9131.

So there is a 1-0.9131 = 0.0869 = 8.69% probability that at least 285 rooms will be occupied.

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Graph A.

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