Question

Five thousand dollars is deposited into a savings account at 2.5% interest compounded continuously.

Answer 1

Answer:

A). A(t) = P(1+r/n)^(nt)

B). DA/Dt = np(1+r/n)^(t)

C). A(5) =$ 5664.0

D).t = approximately 13.5 years

Step-by-step explanation:

A(t) = P(1+r/n)^(nt)

P = $5000

n= t

r= 2.5%

After five years t = 5

A(t) = P(1+r/n)^(nt)

A(5) = 5000(1+0.025/5)^(5*5)

A(5) = 5000(1+0.005)^(25)

A(5)= 5000(1.005)^(25)

A(5) = 5000(1.132795575)

A(5) = 5663.977875

A(5) =$ 5664.0

When the balance A= $7000

A(t) = P(1+r/n)^(nt)

7000= 5000(1+0.025/n)^(nt)

But n= t

7000= 5000(1+0.025/t)^(t²)

7000/5000= (1+0.025/t)^(t²)

1.4= (1+0.025/t)^(t²)

Using trial and error

t = approximately 13.5 years