Question

Calvin deposits $400 in a savings account that accrues 5% interest compounded monthly. After c years, Calvin has

Answer 1

Answer:

Dot number 2. Makayla invests her money 2 years longer.

Step-by-step explanation:

Calvin 400$ with 5% interest compounded monthly will get you to 666.0294029241$ after 13 months or 1 year and 1 month

Makayla 300$ with 6% interest quarterly

1st year 378.743088$(4 quarters)

2nd year 478.1544223592$(+4 quarters)

3rd year 603.6589415506$(+4 quarters)

3rd year and 1 quarter 639.8784780436$

Answer 2

Answer:

So the second option is the answer: Makayla invests her money 2 years longer.

Step-by-step explanation:

Formula for compound interest: $A=P(1+\frac{r}{n})^{nt}$

Calvin:

$A=658.80$

$P=400$

$r=0.05$

$n=12$

$t=?$

Makayla:

$A=613.04$

$P=300$

$r=0.06$

$n=4$

$t=?$

Lets solve for $t$.

1)  Divide both sides of the equation by $P$.

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

2)  Take the natural log of both sides.

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

3)  Rewrite the right side of the equation using properties of exponents.

$ln(\frac{A}{P}) =nt*ln(1+\frac{r}{n})$

4)  Divide each side of the equation by $n*ln(1+\frac{r}{n})$

$\frac{nt*ln(1+\frac{r}n)}{n*ln(1+\frac{r}n)}=\frac{ln(\frac{A}{P})}{n*ln(1+\frac{r}n)}$

5)  Cancel the common factor $n$ on the left side of the equation.

$\frac{t*ln(1+\frac{r}n)}{ln(1+\frac{r}n)}=\frac{ln(\frac{A}{P})}{n*ln(1+\frac{r}n)}$

6)  Cancel the common factor of $ln(1+\frac{r}{n})$.

$t=\frac{ln(\frac{A}{P)}}{n*ln(1+\frac{r}{n})}$

Now we have an equation for $t$ that we can use to answer your question.

For Calvin:

$t=\frac{ln(\frac{658.80}{400)}}{12*ln(1+\frac{0.05}{12})}$

$t=10$

For Makayla:

$t=\frac{ln(\frac{613.04}{300)}}{3*ln(1+\frac{0.06}{3})}$

$t=12$

So the second option is the answer: Makayla invests her money 2 years longer.

Note: This question took me 2 minutes to answer but typing it took 30. lol