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Ahat [919]
3 years ago
9

What would the answer be for this equation 7x/20 - 5x/12

Mathematics
1 answer:
hodyreva [135]3 years ago
4 0

Answer:

\frac{1}{15} x

Step-by-step explanation:

\frac{7x}{20}  -  \frac{5x}{12}

\frac{3 \times 7x}{3 \times 20}  -  \frac{5 \times 5x}{5 \times 12}

\frac{21x}{60}  -  \frac{25x}{60}

\frac{21x - 25x}{60}  = (21 - 25)x =  - 4x

\frac{ - 4x}{60}  =  -  \frac{ x}{15}

\boxed{\green{=  \frac{1}{15} x}}

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Assume that 24.5% of people have sleepwalked. Assume that in a random sample of 1478 adults, 369 have sleepwalked. a. Assuming t
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Answer:

a) 0.3483 = 34.83% probability that 369 or more of the 1478 adults have sleepwalked.

b) 369 < 403.4, which means that 369 is less than 2.5 standard deviations above the mean, and thus, a result of 369 is not significantly high.

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Step-by-step explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

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

The expected value of the binomial distribution is:

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Normal probability distribution

Problems of normally distributed distributions 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)}.

Assume that 24.5% of people have sleepwalked.

This means that p = 0.245

Sample of 1478 adults:

This means that n = 1478

Mean and standard deviation:

\mu = 1478*0.245 = 362.11

\sigma = \sqrt{1478*0.245*0.755} = 16.5346

a. Assuming that the rate of 24.5% is correct, find the probability that 369 or more of the 1478 adults have sleepwalked.

Using continuity correction, this is P(X \geq 369 - 0.5) = P(X \geq 368.5), which is 1 subtracted by the p-value of Z when X = 368.5.

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Z = \frac{368.5 - 362.11}{16.5346}

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c. What does the result suggest about the rate of 24.5%?

Since the sample result is not significant, it suggests that the rate of 24.5% is a good estimate for the percentage of people that have sleepwalked.

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