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Dmitriy789 [7]
3 years ago
12

Which is the value of this expression when p = 3 and q = negative 9? ((p Superscript negative 5 Baseline) (p Superscript negativ

e 4 Baseline) (q cubed)) Superscript 0 Negative one-third Negative StartFraction 1 Over 27 EndFraction StartFraction 1 Over 27 EndFraction One-third
Mathematics
2 answers:
Korvikt [17]3 years ago
8 0

Answer:

The answer is -1/27

Step-by-step explanation:

this is just the person above me's answer except it gets rid of the nonsense words.

pychu [463]3 years ago
7 0

Answer:

Negative StartFraction 1 Over 27 EndFraction

(-1/27)

Step-by-step explanation:

Given expression

(P^-5)(p^-4)(q^3)

When

P=3 and q=-9

(P^-5)(p^-4)(q^3)

=(3^-5)(3^-4)(-9^3)

=(1/3^5)(1/3^4)(-9^3)

=(1/243)(1/81)(-729)

Multiply all numerators

We have,

= (-729)/(243)(81)

Multiply all denominators

= -729/19,683

= -1/27

Negative StartFraction 1 Over 27 EndFraction

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A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben
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Answer:

The estimation for the proportion for this case is \hat p = 0.21

And we know that the 99% confidence interval is given by:

0.21 \pm 0.045

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

The confidence interval would be given by this formula  

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=2.58

The estimation for the proportion for this case is \hat p = 0.21

And we know that the 99% confidence interval is given by:

0.21 \pm 0.045

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

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