Question

I will provide another picture with the questions for this problem provided Please, note that this lengthy. And “complex” if you will, and the subject is pre calculus.

Answer 1

Answer:

• Albert's balance =$2159.07

,

• Marie's balance = $2244.99

,

• Han's balance = $2188.35.

,

• Max's balance = $2147.36

,

• Marie

Explanation:

The compounded interest formula for an amount (Ao) compounded k times in a year over a period of t years at r% per annum is:

$A(t)=A_o(1+\frac{r}{k})^{kt}$

For continous compounding, we make use of the formula:

$A(t)=A_oe^{rt}$

Albert

$1000 earned 1.2% annual interest compounded monthly

$A(10)=1000(1+\frac{1.2}{12\times100})^{12\times10}=\ 1127.43$

$500 lost 2% over the course of the 10 years

$A(10)=500(1-\frac{2}{100})=500\times0.98=\ 490$

$500 grew compounded continuously at rate of $0.8% annually.

$A(10)=500\times e^{0.008\times10}=\ 541.64$

Albert's balance after 10 years will be: 1127.43+490+541.64=$2159.07

Marie

$1500 earned 1.4% annual interest compounded quarterly.

$A(10)=1500(1+\frac{1.4}{4\times100})^{4\times10}=\ 1724.99$

$500 gained 4% over the course of 10 years

$A(10)=500(1+\frac{4}{100})=500\times1.04=\ 520$

Marie's balance after 10 years will be: 724.99+520=$2244.99

Hans

$2000 grew compounded continuously at rate of 0.9% annually.

$A(10)=2000\times e^{\frac{0.9}{100}\times10}=\ 2188.35$

Han's balance after 10 years will be $2188.35.

Max

$1000 decreased in value exponentially at a rate of 0.5% annually.

$A(10)=1000(1-0.5\%)^{10}=1000\times0.995^{10}=\ 951.11$

$1000 earned 1.8% annual interest compounded biannually (twice a year).

$A(10)=1000(1+\frac{1.8}{2\times100})^{2\times10}=\ 1196.25$

Max's balance after 10 years will be 951.11+1196.25=$2147.36

Therefore, after 10 years:

• Albert's balance =$2159.07

,

• Marie's balance = $2244.99

,

• Han's balance = $2188.35.

,

• Max's balance = $2147.36

Since Marie's balance is the highest, she is $10,000 richer.