I would recommend MathPapa for this.
I think it’s 5 square yards I’m not sure though
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
Step-by-step explanation:
In words it is: The product of -11 and k is 22.
The solution is: -11 = +11
22+11 = 33
33 = k
Answer:
9
Step-by-step explanation:
-27/-3
27/3 is 9
each of them have a negative so it stays positive
9
