For large sample confidence intervals about the mean you have:
xBar ± z * sx / sqrt(n)
where xBar is the sample mean z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α sx is the sample standard deviation n is the sample size
We need only to concern ourselves with the error term of the CI, In order to find the sample size needed for a confidence interval of a given size.
z * sx / sqrt(n) = width.
so the z-score for the confidence interval of .98 is the value of z such that 0.01 is in each tail of the distribution. z = 2.326348
The equation we need to solve is:
z * sx / sqrt(n) = width
n = (z * sx / width) ^ 2.
n = ( 2.326348 * 6 / 3 ) ^ 2
n = 21.64758
Since n must be integer valued we need to take the ceiling of this solution.
n = 22
Given sets A = {t, u, v, w, x, y, z} and B = {p, q, r, s, t, u}. Find A and B
alexgriva [62]
Answer:
okay so
so we have A and B = {t,u}
2(10-3) distribute = 2*10 is 20 and 2*3 is 6
20-6+(5-14/2)
now 20-6 is 14
14+(5-14/2)
now 14/2 is 7
14+(5-7)
now 5-7 is -2
14+-2
is 12
the answer is 12
but if you want to you can do 12/3 to get the greatest common factor
which is 4
hope i help you :P
Answer:
25 dfferent ways
Step-by-step explanation:
The answer is 25 different ways because since there are students and thy line up behind each other you would do 5 times 5 which equals to 25.
