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mart [117]
3 years ago
9

PLEASE HELP!!!!!!!!!!!

Mathematics
1 answer:
Anna71 [15]3 years ago
7 0

Answer:

A. X=4

B. X=5

C. N=3

Step-by-step explanation: Hope this help :D

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How are operations with variables like operations with real numbers?
amid [387]

Answer:

The relation between inputs and outputs is given by an equation called the equation of function. Since the outputs are also real values, hence, reasoning by analogy, you may use the arithmetical operation between functions (addition, subtraction, multiplication, division) to produce new functions.

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An article written for a magazine claims that 78% of the magazine's subscribers report eating healthily the previous day. Suppos
Schach [20]

Answer:

89.44% probability that less than 80% of the sample would report eating healthily the previous day

Step-by-step explanation:

We use the binomial approximation to the normal 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:

p = 0.78, n = 675

So

\mu = E(X) = np = 675*0.78 = 526.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{675*0.78*0.22} = 10.76

What is the approximate probability that less than 80% of the sample would report eating healthily the previous day?

This is the pvalue of Z when X = 0.8*675 = 540. So

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

Z = \frac{540 - 526.5}{10.76}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

89.44% probability that less than 80% of the sample would report eating healthily the previous day

8 0
3 years ago
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7nadin3 [17]

It goes like this 1/2, 1/4, 1/8. But if your asking about what fraction didn’t get affected than that just zero because the original peace was cut in half.

If that’s not what you mean can you please elaborate to me!

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