A) Boats depreciate as soon as they are put in the water at rate of 12% yearly. What would be the price (to the nearest penny) of a $29,000 boast after 16 years.

Question

B) $2,000 is invested in a bank that pays an annual interest rate of 5%. At the end of 18 years, how much money will you have if the bank compounds continuously?

Answer 1

Things that grow or shrink by a constant amount are modeled by an exponential function. The generic exponential function is:

where A is the amount at any particular time (t), A₀ is the initial amount (the amount at time t = 0), and "r" is the rate of increase or decrease. Note that you must use the same units for "r" and "t" so if "r" is the rate per year, then "t" must use "years". Also note that YOU must choose whether or not its a plus (+) or a minus (-) depending on whether it's growth (that would be "plus") or decrease (that would be "minus"). It's not both plus and minus like a quadratic or anything.

Since in problem A, we are talking "depreciation", the boat value is going down so we would use a minus. 12% in decimal form is 0.12, t = 16, A₀ = 29,000 so:

A = 29000(1 - 0,12)¹⁶

A = 29000(0.88)¹⁶

A = 29000(0.1293369914)

A = 3750.77

So the boat value is $3,750.77 after 16 years

Problem B uses a different equation. It says the money "compounds continuously" meaning interest is continuously being paid as opposed to being paid once a year. That means you would use this equation:

Again A is the amount at any particular time (t), A₀ is the initial amount (the amount at time t = 0), and "r" is the rate of increase (same unit as "t" and "t" is the time. Note that "e" is a irrational number (about 2.718) sometimes called the "natural number" and it's on your calculator.

So:

A = 2000(e)^0.05(18)

A = 2000(e)^0.9

A = 2000(2.459603111)

A = 4919.21

So after 18 years, you would have $4.919.21 in the bank

Answer 2

The formula is:

P = 29000(1-0.12)^16 = 3750.77 which is the price after 16 years.