In 1850 , a person sold a house to a lady for . If the lady had put the into a bank account paying interest, how much would the investrnent have been w 2010 il interest were compounded in the following ways?
a. mantitly
b. continuously
a. If compounded monthly, the investment would be worth in 2010. (Round to the nearest dollar as needed.)

Ask by Cervantes Smith. in the United States

Mar 23,2025

Question

In 1850 , a person sold a house to a lady for . If the lady had put the into a bank account paying interest, how much would the investrnent have been w 2010 il interest were compounded in the following ways?
a. mantitly
b. continuously
a. If compounded monthly, the investment would be worth in 2010. (Round to the nearest dollar as needed.)

Ask by Cervantes Smith. in the United States

Mar 23,2025

UpStudy AI · Accepted Answer

We are given the following data:
- Principal: $$ P = 30 $$
- Annual rate: $$ r = 0.06 $$ (6%)
- Time: $$ t = 160 $$ years (from 1850 to 2010)
- Compounding frequency: Monthly, so $$ n = 12 $$ times per year

The formula for compound interest is:
$$
A = P\left(1 + \frac{r}{n}\right)^{n t}
$$
Substitute the given values:
$$
A = 30\left(1 + \frac{0.06}{12}\right)^{12 \times 160} = 30\left(1.005\right)^{1920}
$$

Now, we take the natural logarithm of the factor:
$$
\ln\left( (1.005)^{1920} \right) = 1920 \ln(1.005)
$$
For a small rate, we can approximate:
$$
\ln(1.005) \approx 0.0049875
$$
Thus,
$$
1920 \ln(1.005) \approx 1920 \times 0.0049875 \approx 9.577
$$
Convert back from the logarithm:
$$
(1.005)^{1920} \approx e^{9.577}
$$

Calculating $$ e^{9.577} $$:
$$
e^{9.577} \approx 14430 \quad \text{(approximation)}
$$

Thus, the accumulated value is:
$$
A \approx 30 \times 14430 \approx 432900
$$

Rounded to the nearest dollar, if compounded monthly the investment would be worth approximately:
$$
\boxed{\$432{,}900}
$$
If the lady had invested \$30 at 6% interest compounded monthly since 1850, the investment would be worth approximately \$432,900 in 2010.

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