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3 years ago

How do you write 0.512 as a fraction

Mathematics

1 answer:

lesya692 [45]3 years ago

3 0

3 digits after a decimal place = 3 zeros.
0.512×1000=512/100
Answer: 512/1000

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Sylvia has a total of 16 trading cards: 8 superhero cards, 4 princess cards,2 villain cards, and 2 wildlife cards. what is the r

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There as 16 cards, 2 of which are wildlife cards, so the ratio is 16 : 2. We can simplify it by dividing both sides by 2, so we'll get the ratio of 8 : 1.

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4 years ago

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex

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Answer:

The 90% confidence interval that estimate the true proportion of students who receive financial aid is

$0.533 < p < 0.64$

$n = 1789$

Step-by-step explanation:

Considering question a

From the question we are told that

The sample size is n = 200

The number of student that receives financial aid is $k = 118$

Generally the sample proportion is

$\^ p = \frac{k}{n}$

=> $\^ p = \frac{118}{200}$

=> $\^ p = 0.59$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }$

=> $E = 0.057$

Generally 90% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.533 < p < 0.64$

Considering question b

From the question we are told that

The margin of error is E = 0.03

From the question we are told the confidence level is 99% , hence the level of significance is

$\alpha = (100 - 99 ) \%$

=> $\alpha = 0.01$

Generally from the normal distribution table the critical value of is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the sample size is mathematically represented as

$[\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )$

=> $n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )$

=> $n = 1789$

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Step-by-step explanation:

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