Question

PLEASE HELP

Answer 1

Answer:

$r\approx0.03\text{ or about 3\% }$

Step-by-step explanation:

The standard compound interest formula is given by:

$\displaystyle A=P(1+\frac{r}{n})^{nt}$

Where A is the amount afterwards, P is the principal, r is the rate, n is the times compounded per year, and t is the number of years.

Since we are compounding annually, n=1. Therefore:

$\displaystyle A = P (1+r)^{t}$

Lester wants to invest $10,000. So, P=10,000.

He wants to earn $1000 interest. Therefore, our final amount should be 11000. So, A=11000.

And our timeframe is 3.3 years. So, t=3.3. Substituting these values, we get:

$11000=10000(1+r)^{3.3}$

Let’s solve for our rate r.

Divide both sides by 10000:

$1.1=(1+r)^{3.3}$

We can raise both sides to 1/3.3. So:

$\displaystyle (1.1)^\frac{1}{3.3}=((1+r)^{3.3})^\frac{1}{3.3}$

The right side will cancel:

$\displaystyle r+1=(1.1)^\frac{1}{3.3}$

So:

$\displaystyle r=(1.1)^\frac{1}{3.3}-1$

Use a calculator:

$r\approx0.03$

So, the annual rate of interest needs to be about 0.03 or 3% in order for Lester to earn his interest.