Juan invested $3,100 in an account paying an interest rate of 3.6% compounded continuously. Assuming no deposits or withdrawals are made, how much money, to the nearest dollar, would be in the account after 19 years?

Respuesta :

Answer:

A=6143.54( to the nearest dollar A=6144)

Step-by-step explanation:

A=Pe^(r(t))

P=3100 r=0.036 t=19

A=3100e^0.036(19)

A=3100e^(0.684)

There will be $6070 money in the account of Juan after 19 years.

What is Compound Interest?

It is the interest on a loan or deposit calculated based on both the initial principal and accumulated interest from previous periods. It is calculated as:

[tex]Amount=P(1+\frac{r}{n} )^{nt}[/tex]

where,

A = final amount

P = initial principal balance

r = interest rate

n = number of times interest applied per time period

t = number of periods elapsed

We have

principal, P= $3100

Rate, r = 3.6%=[tex]\frac{3.6}{100}[/tex]

time, t = 19years

number of times interest compounded per year, n=1

Formula Used:

[tex]Amount=P(1+\frac{r}{n} )^{nt}[/tex]

Substitute the values in the above formula

[tex]Amount=3100(1+\frac{3.6}{100} )^{19}[/tex]

[tex]Amount=[/tex]$6070

Hence, there will be $6070 in account after 19 years.

Learn more about compound interest here:

https://brainly.com/question/19318210

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