A company rents out 17 food booths and 24 game booths at the county fair. The fee for a food booth is $50 plus $6 per day. The f
1 answer:
Given that 15 food booths were rented out for ‘d’ days at $100 plus $5 per day. On the food booths only, the company will earn 15($100+$5d) = $1500 + $75d Given that 20 game booths were rented out for ‘d’ days at $50 plus $7 per day. On the game booths only, the company will earn 20($50 + $7d) = $1000 + $140d
The total amount that the company is paid is therefore, $1500 + $75d + $1000 + $140d = $2500 + $215d
You might be interested in
Answer:
B. 10000 years old, with a margin of error of 400 years
Step-by-step explanation:
Assuming this complete problem: "12. An archaeologist uses an accelerator mass spectrometer to find the age of a buried branch. At the 68% confidence level, the spectrometer estimates that the branch was 10,000 years old with a following could the spectrometer estimate as the age of the branch at the 95% confidence level?"
The possible options are:
A. 9500 years old, with a margin of error of 500 years
B. 10000 years old, with a margin of error of 400 years
C. 9500 years olds, with a margin of error of 50 years
D. 10000 years old, with a margin of error of 40 year
Solution to the problem
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".
represent the sample mean for the sample
population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
The confidence interval for the mean is given by the following formula:
(1)
The margin of error for this case is given by this formula:
For the first case the confidence was 68% so then and
we can find a critical value on the normal standard distribution that accumulates 0.16 of the area on each tail and this value is
, because P(z<-0.944) = 0.16 and P(Z>.944) = 0.16
The margin of error can be expressed like this:
We can solve for the standard error on this case and we got:
And then for the new confidence interval we need to calculate the new . For the first case the confidence was 95% so then
and
we can find a critical value on the normal standard distribution that accumulates 0.025 of the area on each tail and this value is
, because P(z<-1.96) = 0.025 and P(Z>1.96) = 0.025. So then the new margin of error would be:
The estimation for the mean not changes and from the before and new case is 1000 years. So then the best option for this case would be:
B. 10000 years old, with a margin of error of 400 years
And makes sense since a larger confidence interval means a wider interval.