Answer:0.04
Step-by-step explanation:
For the answer to the question above,
<span>V(n) = a * b^n, where V(n) shows the value of boat after n years.
V(0) = 3500
V(2) = 2000
n = 0
V(0) = a * b^0 = 3500
a = 3500
V(2) = a * b^2
2000 = 3500 * b^2
b = sqrt (2000/3500)
b ≈ 0.76
V(n) = 3500 * 0.76^n
We can check it for n = 1 which is close to 2500 in the graph:
V(1) = 3500 * (0.76)^1
V(1) = 2660
And in the graph we have V(3) ≈ 1500,
V(n) = 3500 * (0.76)^3 ≈ 1536
Now n = 9.5
V(9.5) = 3500 * (0.76)^(9.5)
V(9.5) ≈ 258</span>
Assuming the price of the pool does not change, then
Additional interest required is 4500-3750=750
using
I=Pit, we set up equation
750=3750*0.055*t
Solve for t
t=750/(3750*0.055)=3.64 years.
So will need to wait 3.64 years, or rounded up to 4 years.
