Acompany will need 40,000 in 6 years for a new addition. to meet the goal, the company deposits money into an account today that pays 4% annual intrest compund quarterly.

"A company will need 40,000 in 6 years for a new addition. To meet the goal, the company deposits money into an account today that pays 4% annual intrest compund quarterly."  Let's pretend that the instructions state, "Determine the amount of money that must be deposited upfront so that you will have $40,000 in 6 years."

Use the Compound Amount formula:  A = P(1 + r/n)^(nt),
where P is the principal (the amount deposited upfront), r is the interest rate as a decimal fraction, n is the number of compounding periods, and t is the time in years.

Here, $40000 = P(1 + 0.04/4)^(4*6)
                                                               $40000
So the upfront $ needed is      P =  
                                                           (1+0.01)^24

This comes out to $31502.65 (answer)

Answers:

11) D. $3012) Two answers: choice C, choice D13) A. Subtract 100 from each side14) A. n = 9 and n = -5

Explanation:

11) We set p(120-4p) equal to zero to find that p = 0 is one solution and 120-4p = 0 leads to p = 30. The prices of $0 and $30 both lead to $0 in revenue. It sounds like your teacher wants the larger price value.12) We set each of the two revenue expressions equal to 600.13) Replace the (n-2)^2 with another variable. Let's call that x. So we have the equation 100+x = 149, which is the same as x+100 = 149. We subtract 100 from both sides to undo the +100.14) After following the steps in problem 13, you should have (n-2)^2 = 49, which then becomes n-2 = 7 or n-2 = -7 after applying the square root to both sides. Those two equations solve to n = 9 and n = -5 in that order.