Question

Suppose that $11,399 is invested at an interest rate of 6 3% 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 - 1 (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 $ (Simplify your answers. Round to two decimal places as needed.) The balance after 2 years is $ (Simplify your answers. Round to two decimal places as needed.) The balance after 5 years is $ (Simplify your answers Round to two decimal places as needed.) The balance after 10 years is $. (Simplify your answers. Round to two decimal places as needed.) C) The doubling time is years (Simplify your answers. Round to one decimal place as needed.)

Answer 1

continuously compounded rota Рет A Interest Formula: time in years rate of interest mathematical constant e. and 2.7183 1 Amount Proincipal ent The exponential growth Function 0.063t P(t) = Po e 11,399x where po = initial pricipal amount annual interest rate t length of time the into rest is applied. pct) Future amount at time to b2 Tree balance after year is 11,399 x e E $ 12, 140.24 The balance after 2 years is $ 11,399 xe 0.063 0.06 3X2 12, 929.68 0.063 X 5 The balance after 5 years is $ 11,399 x e $ 15,619.58 The balance after 10 ears 0.06 3x10 is $ 11,399 xe = $ 2,40 2,88 set doubling time bet years then PCt) apo and 2 Po of, nt nt e or, nt ln (2) = line nt 2 t en (2) en (2) The doubling time is on, oro, en (2) 0.063 0.9% = 11 years