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saveliy_v [14]
2 years ago
14

A factory produced 150 tons of steel in one year. the next year, the output increased to 180 tons. what is the percentage increa

se in the output?
Mathematics
2 answers:
Murljashka [212]2 years ago
7 0
<span>A factory produced 150 tons of steel in one year. the next year, the output increased to 180 tons. Then,The percentage increase = (Final -initial) /initial *100 = (180-150)*100/150 = 20% Thus, the net percentage increase in the output is 20%</span>
Fantom [35]2 years ago
6 0

Answer:

20%

Step-by-step explanation:

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0.50x + 7.50 = 0.75(x+6)

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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
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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:

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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.

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