A person places 5800 in an investment account earning an annual rate of 7.9%, compounded continuously. Using the formula V, equals, P, e, start superscript, r, t, end superscriptV=Pe^{rt}, where V is the value of the account in t years, P is the principal initially invested, e is the base of a natural logarithm, and r is the rate of interest, determine the amount of money, to the nearest cent, in the account after 11 years.

Solution

A p erso n pl a ces 5800 in an in v es t m e n t a cco u n t e a r nin g an ann u a l r a t e o f 7.9%, co m p o u n d e d co n t in u o u s l y . U s in g t h e f or m u l a V, e q u a l s, P, e, s t a r t s u p erscr i pt, r, t, e n d s u p erscr i pt V = P e^{r t}, w h ere V i s t h e v a l u e o f t h e a cco u n t in t ye a rs, P i s t h e p r in c i p a l ini t ia ll y in v es t e d, e i s t h e ba se o f a na t u r a l l o g a r i t hm, an d r i s t h e r a t e o f in t eres t, d e t er min e t h e am o u n t o f m o n ey, t o t h e n e a res t ce n t, in t h e a cco u n t a f t er 11 ye a rs .

Amount : 13830.25

étapes des solutions

Translate the problem into an equation:

A = 5800 e^{0.079 \cdot 11}

5800 e^{0.079 \cdot 11} = 13830.24578 \dots

Arrondir à hundredth la plus proche:

13830.25

Amount : 13830.25