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rewona [7]
3 years ago
14

A train travels 120 km in 2 hours and 30 miniutes. what is the average speed in km/h ?​

Mathematics
1 answer:
elena-s [515]3 years ago
6 0

Answer:

120/2.5 = 48

48 km/hr

Step-by-step explanation:

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a line passes through (2, –1) and (8, 4). a. choose the correct equation for the line in point-slope form. b. what is the equati
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In standard form, 6y + 6 = 5x - 10 => 5x - 6y = 6 + 10 => 5x - 6y = 16
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(03.01) Is the following relation a function? y equals the sine of 3 x
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Which is the graph of f(c)=x^2-2x+3
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Suppose that 20% of the adult women in the United States dye or highlight their hair. We would like to know the probability that
Rasek [7]

Answer:

71.08% probability that pˆ takes a value between 0.17 and 0.23.

Step-by-step explanation:

We use the binomial approxiation 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 = 0.2, n = 200. So

\mu = E(X) = np = 200*0.2 = 40

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.66

In other words, find probability that pˆ takes a value between 0.17 and 0.23.

This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So

X = 46

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

Z = \frac{46 - 40}{5.66}

Z = 1.06

Z = 1.06 has a pvalue of 0.8554

X = 34

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

Z = \frac{34 - 40}{5.66}

Z = -1.06

Z = -1.06 has a pvalue of 0.1446

0.8554 - 0.1446 = 0.7108

71.08% probability that pˆ takes a value between 0.17 and 0.23.

6 0
3 years ago
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