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arlik [135]
3 years ago
11

A video rental store has 6,000 movies. One Friday, 3/5 of the movies were rented. How many movies were rented that Friday night?

Mathematics
2 answers:
AleksAgata [21]3 years ago
8 0
6000/ 5 = 1200 x 3 =3600 movies
vazorg [7]3 years ago
4 0
Explanation:

<span><span>35</span>=<span>610</span></span>

<span>XX=<span>6001000</span></span>

<span>XX=<span>36006000</span></span>

That is, if 3 out of 5 movies were rented
then this is the same as 3600 out of 6000 movies were rented.

there were 3600 movies rented out that evening.

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5x+3y=9 in standard form
tamaranim1 [39]
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The probability of flu symptoms for a person not receiving any treatment is 0.038. In a clinical trial of a common drug used to
alexgriva [62]

Answer:

36.32% probability that at least 47 people experience flu symptoms. This is not an unlikely event, so this suggests that flu symptoms are not an adverse reaction to the drug.

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

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:

n = 1164, p = 0.038

So

\mu = E(X) = np = 1164*0.038 = 44.232

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = 6.5231

Estimate the probability that at least 47 people experience flu symptoms.

Using continuity correction, this is P(X \geq 47 - 0.5) = P(X \geq 46.5), which is 1 subtracted by the pvalue of Z when X = 46.5.

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

Z = \frac{46.5 - 44.232}{6.5231}

Z = 0.35

Z = 0.35 has a 0.6368

1 - 0.6368 = 0.3632

36.32% probability that at least 47 people experience flu symptoms. This is not an unlikely event, so this suggests that flu symptoms are not an adverse reaction to the drug.

6 0
3 years ago
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