Answer:
Step-by-step explanation:
Given that
Group Group One Group Two
Mean 26.00 23.00
SD 2.00 4.00
SEM 0.40 1.21
N 25 11
where group I represents female servers and group II male servers.
We have to calculate confidence interval for 90% for difference in means
The mean of Group One minus Group Two equals 3.00
df = 34
standard error of difference = 0.993
t critical = 2.034 for 90% df 34
Hence confid. interval at 90%
=Mean diff ±2.034 * std error of diff
= (0.98, 5.02)
Answer:
5x^2 + 13x
Step-by-step explanation:
If nCk represents the number of ways k parts can be chosen from a pool of n, the probability of interest is the complement of the probability of selecting all good parts.
1 - (167C3)/(170C3) = 42,085/804,440 ≈ 0.0523
_____
nCk = n!/(k!(n-k)!)
0.3 if a higher number because anytime there is a negative that means the number is below zero. 0.3 is barley above 0. But -0.7 is below zero
