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)
Answer:
x^5
Step-by-step explanation:
First factor -12m^n - 49mn - 44n^2 to get -(4n+3m)(11n + 4m) then the equation would be:
-(3m + 4n)(4m + 11n) / (-3m - 4n)
Then, cancel out the like terms and the final answer would be
4m + 11n
Answer:
270 dollars for a year, so 22.50 times 12
Since we know the value for c and d we just plug them into the equation which will look like this. 6•5^2-5•4+8
Answer: 138
