A person invests 10000 dollars in a bank. The bank pays 6.5% interest compounded annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 29500 dollars?

Lösung

A p erso n in v es t s 10000 d o ll a rs in a bank . T h e bank p a ys 6.5% in t eres t co m p o u n d e d ann u a ll 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 29500 d o ll a rs ?

Time : 17.18

Schritte zur Lösung

Translate the problem into an equation:

29500 = 10000 (1 + \frac{0.065}{1})^{1 \cdot t}

Löse nach

t, 29500 = 10000 (1 + \frac{0.065}{1})^{1 \cdot t} : t = 17.18

Time : 17.18