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Ganezh [65]
3 years ago
6

At a carnival, Ann sold 314 bottles of drinks a day. She sold 66 bottles more in the afternoon then in the morning. How many bot

tles of drinks did she sell in the morning?
Mathematics
1 answer:
Paladinen [302]3 years ago
8 0

Answer:

Ann sold 124 bottles of drinks in the morning

Step-by-step explanation:

Let

x ---> bottles sold in the afternoon

y ---> bottles sold in the morning

we know that

Ann sold 314 bottles of drinks a day

so

x+y=314 -----> equation A

She sold 66 bottles more in the afternoon than in the morning

so

x=y+66 ----> equation B

substitute equation B in equation A

y+66+y=314

solve for y

2y=314-66\\2y=248\\y=124

therefore

Ann sold 124 bottles of drinks in the morning

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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
work for a publishing company. The company wants to send two employees to a statistics conference. To be​ fair, the company deci
NISA [10]

Answer:

Step-by-step explanation:

Here is the complete question.

Dominique, Marco, Roberto , and John work for a publishing company. The company wants to send two employees to a statistics conference. To be fair, the company decides that the two individuals who get to attend will have their names drawn from a hat. This is like obtaining a simple random sample of size 2. (a) Determine the sample space of the experiment. That is, list all possible simple random samples of size n = 2. (b) What is the probability that Dominique and Marco attend the conference? (c) What is the probability that John attends the conference?  ​(d) What is the probability that John stays​ home?

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Let D = Dominique, M = Marco, R = Roberto and J = John

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b) Probability is the ratio of number of event to the sample space.

P = n(E)/n(S)

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n(E) is the event of Dominique and Marco attending the conference.

E = {DM}

n(E) = 1

P(D and M) = 1/6

Hence  the probability that Dominique and Marco attend the conference is 1/6

c) For John to attend the conference, the event outcome will be given as;

E = {DJ, MJ, RJ}

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d) Probability that John stays at home = 1 - Prob (John attends the conference)

Probability that John stays at home = 1 - 1/2

Probability that John stays at home = 1/2

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natima [27]
First we rewrite the table:
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