Answer:
18.83% (nearest hundredth)
Step-by-step explanation:
Continuous Compounding Formula
[tex]\sf A=P(e)^{rt}[/tex]
where:
- A = amount
- P = principal (initial amount)
- e = mathematical constant ≈ 2.7183
- r = interest rate (in decimal form)
- t = time in years
Given:
- A = $1,300
- P = $420
- t = 6 years
Substituting given values into the formula:
[tex]\sf \implies 1300=420(e)^{6r}[/tex]
[tex]\sf \implies \dfrac{1300}{420}=e^{6r}[/tex]
[tex]\sf \implies e^{6r}=\dfrac{65}{21}[/tex]
Taking natural logs of both sides:
[tex]\sf \implies \ln e^{6r}=\ln \dfrac{65}{21}[/tex]
[tex]\sf \implies 6r\ln e=\ln \dfrac{65}{21}[/tex]
[tex]\sf \implies 6r(1)=\ln \dfrac{65}{21}[/tex]
[tex]\sf \implies 6r=\ln \dfrac{65}{21}[/tex]
[tex]\sf \implies r=\dfrac16 \ln \dfrac{65}{21}[/tex]
[tex]\sf \implies r=0.1883108054...[/tex]
Therefore, the interest rate is 18.83% (nearest hundredth)
