Answer:
1 and 2 go in the 2nd area and 3 gos to the 1st and 4th gos to the last one
Step-by-step explanation:
hope this helped!
P.S. can i pls have brainlyist am trying to lv up thx! ;D
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
Answer:
This is my solution for your question
Answer:
= 7.071067812
Step-by-step explanation:
hope this helps
It is 46 if you divide 34 by 16 you get 2.125 then divide 100 by 2.125 and you get 47.05 the 0 means go down if you are rounding which brings it to 46
