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