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uysha [10]
3 years ago
13

Which is the equation of a line that has a slope of į and passes through point (2, -3)?

Mathematics
1 answer:
Hoochie [10]3 years ago
4 0

Answer:

o Y =1/x+2

please mark as brainliest

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Simplify the following expression. 2x^5+3x^3-5x^2+7x+1+7x^5-3x^3-4
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Answer:

9x^5−5x^2+7x−3

Step-by-step explanation:

2x5+3x3+−5x2+7x+1+7x5+−3x3+−4

(2x5+7x5)+(3x3+−3x3)+(−5x2)+(7x)+(1+−4)

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Step-by-step explanation:

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Please explain 22 and 23
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Hey there!

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Read 2 more answers
A courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the
Orlov [11]

Answer:

95.44% probability that the sample proportion will differ from the population proportion by less than 0.03.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question:

p = 0.05, n = 212, \mu = 0.05, s = \sqrt{\frac{0.05*0.95}{212}} = 0.015

What is the probability that the sample proportion will differ from the population proportion by less than 0.03?

This is the pvalue of Z when X = 0.03 + 0.05 = 0.08 subtracted by the pvalue of Z when X = 0.05 - 0.03 = 0.02. So

X = 0.08

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

By the Central Limit Theorem

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

Z = \frac{0.08 - 0.05}{0.015}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 0.02

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

Z = \frac{0.02 - 0.05}{0.015}

Z = -2

Z = -2 has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

95.44% probability that the sample proportion will differ from the population proportion by less than 0.03.

6 0
3 years ago
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