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Virty [35]
3 years ago
9

Which option or options is correct?

Mathematics
1 answer:
WITCHER [35]3 years ago
6 0

All of them are true.

A: Rule.

B: There exist such functions f and g that satisfy the equality.

C: According to (A) this is acceptable.

D: Rule.

Hope this helps.

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-7 2/3 + (-5 1/2) + 8 3/4
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Step-by-step explanation:

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Jorge has 105 markers. 68 of the markers were dried and thrown out. How many working markers does jorge now have?
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Answer:43

Step-by-step explanation:

you have 105 to start off with

But then 68 of them get dried and thrown out so you stay with

105-68=43


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Step-by-step explanation:

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What is the equation of the following line written in slope-intercept form?
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In a large company, the proportion of employees who were promoted during the last year was 0.10. If 100
Natasha2012 [34]

Answer:

0.0475

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 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)}.

In a large company, the proportion of employees who were promoted during the last year was 0.10.

This means that p = 0.1

100 employees

This means that n = 100

Mean and standard deviation:

\mu = E(X) = np = 100*0.1 = 10

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.1*0.9} = 3

What is the probability that at least 15 of them were promoted during the last year?

This is P(X \geq 15), which is 1 subtracted by the pvalue of Z when X = 15. So

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

Z = \frac{15 - 10}{3}

Z = 1.67

Z = 1.67 has a pvalue of 0.9525

1 - 0.9525 = 0.0475.

0.0475 is the answer.

6 0
3 years ago
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