Question

a bank looks to attract investors. they are offering 9.5% annual interest rate, compounded continuously, on any investment over $12,000. if an investor has $15,500 to invest with the bank, how much would they have in their account after 12 years? round your answer to the nearest cent (hundredth).

Explanation:

Step1: Recall the formula for continuous compounding

The formula for continuous compounding is $A = Pe^{rt}$, where $A$ is the amount of money accumulated after $t$ years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (in decimal), and $t$ is the time the money is invested for in years.

Step2: Identify the values of \(P\), \(r\), and \(t\)

Given that \(P=\$15500\), \(r = 9.5\%=0.095\) (converted from percentage to decimal by dividing by 100), and \(t = 12\) years.

Step3: Substitute the values into the formula

Substitute \(P = 15500\), \(r=0.095\), and \(t = 12\) into the formula \(A=Pe^{rt}\). So we have \(A=15500\times e^{0.095\times12}\).
First, calculate the exponent: \(0.095\times12 = 1.14\). Then the formula becomes \(A = 15500\times e^{1.14}\).
We know that \(e^{1.14}\approx3.1267\) (using a calculator to find the value of the exponential function).
Then \(A=15500\times3.1267\).
Calculate \(15500\times3.1267 = 15500\times3+15500\times0.1267=46500 + 1963.85=48463.85\) (approximate calculation, more accurately using calculator: \(15500\times3.1267 = 15500\times3.1267 = 48463.85\) when calculated precisely).

Answer:

\(\$48463.85\) (Note: If we calculate more precisely, \(e^{1.14}\approx e^{1.14}\approx3.126703744\), then \(15500\times3.126703744 = 15500\times3.126703744=48463.908032\approx48463.91\). There might be a slight difference due to the precision of \(e^{1.14}\) value. If we use a more accurate value of \(e^{1.14}\), the answer is approximately \(\$48463.91\). But based on the step - by - step above with \(e^{1.14}\approx3.1267\), we get \(\$48463.85\). The correct answer with more precise calculation of \(e^{1.14}\) is \(\$48463.91\))

Wait, let's recalculate with more precision. Let's use a calculator for \(e^{1.14}\):

Using a calculator, \(e^{1.14}\):

We know that the Taylor series of \(e^x=\sum_{n = 0}^{\infty}\frac{x^n}{n!}\), but it's easier to use a calculator. \(e^{1.14}\approx3.126703744\)

Then \(A = 15500\times3.126703744=15500\times3.126703744\)

\(15500\times3.126703744=(15000 + 500)\times3.126703744=15000\times3.126703744+500\times3.126703744\)

\(15000\times3.126703744 = 46900.55616\)

\(500\times3.126703744=1563.351872\)

Adding them together: \(46900.55616+1563.351872 = 48463.908032\approx48463.91\)

So the more accurate answer is \(\$48463.91\))