A person invests 3500 dollars in a bank. The bank pays 4.5% interest compounded monthly. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 6900 dollars?

Solution

A p erso n in v es t s 3500 d o ll a rs in a bank . T h e bank p a ys 4.5% in t eres t co m p o u n d e d m o n t h l y . T o t h e n e a res t t e n t h o f a ye a r, h o w l o n g m u s t t h e p erso n l e a v e t h e m o n ey in t h e bank u n t i l i t re a c h es 6900 d o ll a rs ?

Time : 15.11

étapes des solutions

Translate the problem into an equation:

6900 = 3500 (1 + \frac{0.045}{12})^{12 t}

Solutionner pour

t, 6900 = 3500 (1 + \frac{0.045}{12})^{12 t} : t = 15.11

Time : 15.11