Students sold raffle tickets to raise money for a field trip. The first 20 tickets cost $4 each. To sell more tickets, they lowe
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Answer:
The 99% confidence interval would be given (11.448;14.152).
Step-by-step explanation:
1) Important concepts and notation
A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"
represent the sample deviation
represent the sample mean
n =1003 is the sample size selected
Confidence =99% or 0.99
represent the significance level.
2) Solution to the problem
The confidence interval for the mean would be given by this formula
We can use a z quantile instead of t since the sample size is large enough.
For the 99% confidence interval the value of and
, with that value we can find the quantile required for the interval in the normal standard distribution.
And replacing into the confidence interval formula we got:
And the 99% confidence interval would be given (11.448;14.152).
We are confident that about 11 to 14 are the number of books that the people had read on the last year on average, at 1% of significance.
Correct question with correct options has been attached
Answer:
37
Step-by-step explanation:
Let C represent those dressed as heroes and let B represent those wearing mask
Thus;
n(C) = 39
n(D) = 45
We are told that 21 entrants were dressed as heroes wearing a mask.
Thus; n(C n D) = 21
We are also told that there were a total of 100 entries.
Thus, n(T) = 100
Now, number of entrants neither dressed as heroes nor wearing a mask is given by;
n(T) - n(C u D)
n(C u D) = n(C) + n(D) - n(C n D)
n(C u D) = 39 + 45 - 21
n(C u D) = 63
Thus;number of entrants neither dressed as heroes nor wearing a mask is: 100 - 63 = 37
