A high-yield savings account that compounds interest continuously was opened in 1991. The recursive equation f(x)≈f(x−1)⋅1.375

shows the amount of money in the account rounded to the nearest cent for each year after 2000, in which f(0)=3,500
. Use the recursive formula to create an exponential equation that models the account’s balance, A
, after t years, assuming there were no withdrawals or deposits. What is the correct equation, with r rounded to three decimal places and P
rounded to two decimal places? While solving, round values to three decimal places. (1 point)
Responses

A=3,500e0.375t
upper A equals 3,500 e superscript 0.375 t baseline

A=3,500(1.375)t
upper A equals 3,500 left parenthesis 1.375 right parenthesis superscript t baseline

A=3,500e0.318t
upper A equals 3,500 e superscript 0.318 t baseline

A=3,500(1.318)t

1 answer

To develop the exponential equation that models the account's balance \( A \) after \( t \) years, we can start from the recursive formula provided:

\[ f(x) \approx f(x−1) \cdot 1.375 \text{ where } f(0) = 3500 \]

From the recursive relationship, we can deduce how the value of the account changes annually. This means that for each year \( t \), the value can be written as:

\[ f(t) = f(t-1) \cdot 1.375 \]

After \( t \) years, we can unfold this to:

\[ f(t) = 3500 \cdot (1.375)^t \]

This is the standard form of exponential growth where

  • the initial amount (or principal) \( P = 3500 \)
  • the growth factor per year \( r = 1.375 \)

Thus, the equation that models the account's balance after \( t \) years is:

\[ A = 3500 \cdot (1.375)^t \]

So, the correct equation is:

\[ A = 3,500 (1.375)^t \]

Where \( P = 3500 \) (rounded to two decimal places) and \( r = 1.375 \) (which is already rounded to three decimal places).

So the final answer is:

A = 3,500(1.375)^{t}