Given:
n = 50, sample size
, sample mean
s = 2.4 min, sample standard deviation.
The confidence interval is
At the 99% confidence level, t* (from the student's t-distribution) is
t* = 2.68
Therefore
t*(s/√n) = 2.68*(2.4/√50) = 0.9096
The confidence interval is
(23.6-0.9096, 23.6+0.9096) ≈ (22.69, 24.51)
Answer: (22.7, 24.5)
78 is bigger since it's a whole number. 0.78 is less than 1 itself. Hope that I helped :)<span />
Answer:
hehe
Step-by-step explanation:
In expanded form :
100,000+10,000+4,000+6
I don’t know the answer I’m sorry