Answer:
The marginal error is 2.355
Confidence interval is 
The LB and UB are 34.645 and 39.3555 respectively.
Step-by-step explanation:
Consider the provided information.
n = 49, s = 8.2 c = 95%and 
Degree of freedom is: n-1 = 49-1 = 48.

From the table 
Marginal error is:





Hence, the marginal error is 2.355.
Part (B)
95% confidence interval estimate



Thus, the LB and UB are 34.645 and 39.3555 respectively.
The Answer: A = $ 13,421.56
Formula A=P(1 + r/n)^(nt) A=10000(1+0.059/12)^(12(5)) = $ 13,421.56
V = the future value of the investmentP = the principal investment amountr = the annual interest raten = the number of times that interest is compounded per yeart = the number of years the money is invested for
The answer is 7x + 3
First, you have 3x-1 + 2x+1
Combine like terms and the 1’s cancel each other out. You are left with 5x.
Then, you add x+2 + x+1
Combine like terms and you have 2x+3
Add 5x + 2x+3 and you have 7x +3
I’m bad at explaining, so here are the steps.