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lozanna [386]
3 years ago
15

Ivan and Jeff buy a package of 8 pens for $4.00. Ivan wants 5 of the pens, and Jeff wants 3. How much should each student pay?

Mathematics
1 answer:
irga5000 [103]3 years ago
3 0
$4 / 8 = 50 cents per pen

.5 * 5 = 2.50
.5 * 3 = 1.50

Ivan pays $2.50 and Jeff pays $1.50
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Answer:

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A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben
juin [17]

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:

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%)

5 0
3 years ago
A whole number is squared. The result is between 200 and 260. The number is between:
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Answer:

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7 0
2 years ago
Write an equation of the line passing through point p that is perpendicular to the given line. p(3,1), y=13x−5
Anarel [89]
You know two lines y=ax+b and y=mx+n that are perpendicular, so we have the product a*m=-1

+ This line is perpendicular to y=13x-5, so it has equation: y=-1/13x+b
+ And it passes through the point (3;1), so we have x=3, y=1. So 1=-1/13 *3+b
and b= 1+3/13= 16/13
And we have y=-1/13x+16/13
Have fun
7 0
3 years ago
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