Question

(1 point) Redo problem 15 in section 6.3 of your textbook (page 289). Suppose that the machinery in question costs $119000 and earns profit at a continuous rate of $67000 per year. Use an interest rate of 7% per year, compounded continuously. When is the present value of the profit equal to the cost of the machinery? Round your answer to the nearest tenth of a year.

Answer 1

Time will be 8.20 year

Step-by-step explanation:

We have given that the machinery cost A = $11900

And principal amount P = $67000

Rate of interest r = 7 % = 0.07

We have to find the time t

We know that formula

`A=Pe^(rt)`

`119000=67000e^(0.07t)`

`1.776=e^(0.07t)`

`0.07t=ln1.776`

`0.07t=0.5743`

t = 8.20 year