how much more would you earn in the first investment than in the second investment? $49,000 invested for 30…

Explanation:

Step1: Recall compound - interest formula

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, where $A$ is the amount of money accumulated after $n$ years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (in decimal form), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested for in years.

For the first investment: $P = 49000$, $r_1=0.07$, $n_1 = 12$ (compounded monthly), and $t = 30$. $A_1=49000(1 +\frac{0.07}{12})^{12\times30}$ $A_1=49000(1+\frac{0.07}{12})^{360}$ First, calculate $(1+\frac{0.07}{12})\approx1 + 0.005833=1.005833$. Then, $(1.005833)^{360}\approx7.9188$. So, $A_1=49000\times7.9188 = 387021.2$.

Step2: Calculate the amount for the second investment

For the second investment: $P = 49000$, $r_2 = 0.14$, $n_2=1$ (compounded annually), and $t = 30$. $A_2=49000(1 + 0.14)^{30}$ $(1 + 0.14)^{30}\approx267.8635$. So, $A_2=49000\times267.8635=13125311.5$.

Step3: Find the difference

The difference $\Delta A=A_2 - A_1$. $\Delta A=13125311.5-387021.2 = 12738290.3$.

Answer:

$12738290.3$