Solution :
Given :
Sample mean, 
Sample size, n = 129
Sample standard deviation, s = 8.2
a. Since the population standard deviation is unknown, therefore, we use the t-distribution.
b. Now for 95% confidence level,
α = 0.05, α/2 = 0.025
From the t tables, T.INV.2T(α, degree of freedom), we find the t value as
t =T.INV.2T(0.05, 128) = 2.34
Taking the positive value of t, we get
Confidence interval is ,
(32.52, 35.8)
95% confidence interval is (32.52, 35.8)
So with 
confidence of the population of the mean number of the pounds per person per week is between 32.52 pounds and 35.8 pounds.
c. About of confidence intervals which contains the true population of mean number of the pounds of the trash that is generated per person per week and about 
that doe not contain the true population of mean number of the pounds of trashes generated by per person per week.
Answer:
Step-by-step explanation:
18x² + 2 = 2 * 9x² + 2 *1
= 2*(9x² + 1)
Total balance in Mr. Copley at the end of the month = $1473.61.
Initial deposite = $75 (not shown in his checkbook).
Let us assume his balance at the beginning of the month = x.
Total balance at the end of the month = The balance at the beginning of the month + total deposite.
$1473.61 = x + 75.
Subtracting 75 from both sides, we get
1473.61 -75 = x + 75 -75.
1398.61 = x.
Therefore, his balance at the beginning of the month was $1398.61.
B, because nothing repeats.
