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poizon [28]
3 years ago
13

What is the sum? 2/x^2+4/x2

Mathematics
1 answer:
Mashutka [201]3 years ago
8 0
2/x2 +4. So write it as 2+4/x2 your answer should be 6/x2
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A BYU-Idaho professor took a survey of his classes and found that 82 out of 90 people who had served a mission had personally me
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Answer:

The 99% confidence interval for the difference in the two proportions is (-0.0247, 0.2833).

Step-by-step explanation:

Before building the confidence interval, we need to understand the Central Limit Theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes 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}}

A BYU-Idaho professor took a survey of his classes and found that 82 out of 90 people who had served a mission had personally met a member of the quorum of the twelve apostles.

This means that:

p_S = \frac{82}{90} = 0.9111

s_S = \sqrt{\frac{0.9111*0.0888}{90}} = 0.045

Of the non-returned missionaries that were surveyed 86 of 110 had personally met a member of the quorum of the twelve apostles.

This means that:

p_N = \frac{86}{110} = 0.7818

s_N = \sqrt{\frac{0.7818*0.2182}{110}} = 0.0394

Distribution of the difference:

p = p_S - p_N = 0.9111 - 0.7818 = 0.1293

s = \sqrt{s_S^2+s_N^2} = \sqrt{0.045^2+0.0394^2} = 0.0598

Calculate a 99% confidence interval for the difference in the two proportions.

The confidence interval is:

p \pm zs

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

99% confidence level

So \alpha = 0.01, z is the value of Z that has a p-value of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

The lower bound of the interval is:

p - zs = 0.1293 - 2.575*0.0598 = -0.0247

p + zs = 0.1293 + 2.575*0.0598 = 0.2833

The 99% confidence interval for the difference in the two proportions is (-0.0247, 0.2833).

5 0
2 years ago
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