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
Well the increase is 132/120. Simplify by dividing 12 from the numerator and the denominator, you should get 11/10.
11/10 is also 110%
Answer:
a^2/3
Step-by-step explanation:
the numerator simplifies to 5a^3 and the bottom is 15a. we can use our exponent rules to subtract exponents, so it is 1/3(a^2). this simplifies to a^2/3.
C. 118.2
1 and 2 make a complimentary angle, so 2=118.2
2 and 6 are corresponding, so angle 2=angle 6 (118.2 = 118.2)