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:
what do you need help with? i can edit this later lol
1 ) 9 x + 18 > 9 x - 27
9 x - 9 x > - 27 - 18
0 * x > - 45
Always true
2 ) 6 x - 13 < 9 x - 12
6 x - 6 x < - 12 + 13
0 * x < 1
Always true
3 ) - 6 ( 2 x - 10 ) + 12 x ≤ 180
- 12 x + 60 + 12 x ≤ 180
0 * x ≤ 180 - 60
0 * x ≤ 120
Always true.
Answer:
I think 5
Step-by-step explanation:
9 games
You can get this by multiplying each percentage by the total number of games to see what they have to expect.
.55*180 = 99 games (red team)
.60*180 = 108 games (blue team)
Now subtract the blue team expectation from the red team.
108 - 99 = 9 games.