Studdy AI Logo
Math  /  Algebra

QuestionSuppose that $12,251\$ 12,251 is invested at an interest rate of 6.6%6.6 \% per year, compounded continuously. a) Find the exponential function that describes the amount in the account after time t , in years. b) What is the balance after 1 year? 2 years? 5 years? 10 years? c) What is the doubling time? a) The exponential growth function is P(t)=12251e0.066t\mathrm{P}(\mathrm{t})=12251 e^{0.066 t} (Type exponential notation with positive exponents. Do not simplify. Use integers or decimals for any numbers in the equation.) b) The balance after 1 year is $\$ □\square (Simplify your answers. Round to two decimal places as needed.)

Studdy Solution
Compute the doubling time:
t=ln⁡(2)0.066≈0.69310.066≈10.50 t = \frac{\ln(2)}{0.066} \approx \frac{0.6931}{0.066} \approx 10.50
The doubling time is approximately 10.50 years.
The exponential function is P(t)=12251e0.066t P(t) = 12251 e^{0.066t} . The balance after 1 year is approximately \$13,086.56. The balance after 2 years is approximately \$13,990.79. The balance after 5 years is approximately \$17,043.06. The balance after 10 years is approximately \$23,717.88. The doubling time is approximately 10.50 years.

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord