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2 years ago

Which of the following is a better buy? 7 tickets for $22.75 10 tickets for $31.50

Mathematics

2 answers:

AleksAgata [21]2 years ago

8 0

Answer:

10 tickets for $31.50

Step-by-step explanation:

So If we divide 22.75 by 7 it gives you 3 dollars and 25 cents per ticket

If you divide 31.50 by 10 it gives you 3 dollars and 15 cents.

So In conclusion 10 tickets for $31.50 dollars is the better buy

kkurt [141]2 years ago

4 0

Answer:

10 Tickets Is The Better Deal

Step-by-step explanation:

So What You Do Is You Will Divide 22.75 By 7

You Will Get 3.25! That Means Its 3.25 Each For 7 Tickets

For The Next One You Do The Same Thing

You Want To Divide 31.50 By 10

You Will Get 3.15! That Means Its 3.15 Each For 10 Tickets

So 3.15 Is 10 Cheaper Than 3.25 So Getting 10 Tickets Is A Better Deal

Hope This Help

DOH! My Brain Hurts

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2 years ago

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.