Given:
n = 20, sample size
xbar = 17.5, sample mean
s = 3.8, sample standard deiation
99% confidence interval
The degrees of freedom is
df = n-1 = 19
We do not know the population standard deviation, so we should determine t* that corresponds to df = 19.
From a one-tailed distribution, 99% CI means using a p-value of 0.005.
Obtain
t* = 2.8609.
The 99% confidence interval is
xbar +/- t*(s/√n)
t*(s/√n) = 2.8609*(3.8/√20) = 2.4309
The 99% confidence interval is
(17.5 - 2.4309, 17.5 + 2.4309) = (15.069, 19.931)
Answer: The 99% confidence interval is (15.07, 19.93)
Option C is your answer.
9x² + 6x + 1
9x² + 3x + 3x + 1
3x(3x + 1) + 1(3x + 1)
(3x + 1)(3x + 1)
The third one is correct so it would be C.
Just substitute X and Y into the question.
What does that even mean? can you re-word the question please?
