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andriy [413]
3 years ago
10

______ is the difference between the amount a company earns and the amount of money it spends

Mathematics
1 answer:
sasho [114]3 years ago
3 0
Profit is the difference between a companies revenue and its expenses.
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Substitution Method <br> y=x-4<br> y=4x-10
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x=2 and y= -2

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Substitute y=x-4 in second equation

x-4=4x-10

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3x=6

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If SALARY(HOURS) is the function that describes Al's salary at Bit Labs based on his training, SALARY(56 HOURS), SALARY(56), and
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3 years ago
The mean points obtained in an aptitude examination is 159 points with a standard deviation of 13 points. What is the probabilit
Korolek [52]

Answer:

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 159, \sigma = 13, n = 60, s = \frac{13}{\sqrt{60}} = 1.68

What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?

This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So

X = 160

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

By the Central Limit Theorem

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

Z = \frac{160 - 159}{1.68}

Z = 0.6

Z = 0.6 has a pvalue of 0.7257

X = 150

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

Z = \frac{158 - 159}{1.68}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

7 0
3 years ago
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