refer to the following scenario to answer questions #4 - 5: lucy opened a savings account in the year 2012…

Explanation:

Step1: Identify the regression model

Since there is a fixed - interest rate and no new deposits or withdrawals, this is an exponential growth model of the form $A = P(1 + r)^t$, where $A$ is the account balance, $P$ is the initial deposit, $r$ is the annual interest rate, and $t$ is the number of years. Let $t = 0$ correspond to 2022, so for 2022, $A_1=P(1 + r)^0=P$, for 2023 ($t = 1$), $A_2=P(1 + r)^1$, for 2024 ($t = 2$), $A_3=P(1 + r)^2$, and for 2025 ($t = 3$), $A_4=P(1 + r)^3$. We can find the common ratio $1 + r$ by taking the ratio of consecutive balances. For example, $1 + r=\frac{A_{n + 1}}{A_n}$. Let's take $\frac{1556.38}{1488.08}\approx1.046$.

Step2: Set up the exponential equation

We know that the general form of the exponential equation is $A = P(1.046)^t$. Using the data point for 2022 ($t = 0$, $A = 1488.08$), we can work backward to the initial - deposit year. The number of years from 2012 to 2022 is $t = 10$. So, if $A = P(1.046)^t$, then $P=\frac{A}{(1.046)^t}$. Substituting $A = 1488.08$ and $t = 10$ into the formula, we get $P=\frac{1488.08}{(1.046)^{10}}$. Calculate $(1.046)^{10}\approx1.599$. Then $P=\frac{1488.08}{1.599}\approx930\approx925$ (due to rounding in the process).

Step3: Solve for the account balance in 20 years

Lucy opened the account in 2012. 20 years after 2012 is 2032. From 2022 ($t = 0$) to 2032, $t = 10$. The exponential equation is $A = P(1.046)^t$. We found $P\approx925$, and $t = 10$. $A=925\times(1.046)^{10}$. Since $(1.046)^{10}\approx1.599$, then $A = 925\times1.599 = 925\times(1 + 0.599)=925+925\times0.599=925+553.075 = 2282.4$.

Answer:

4. B. $925
5. B. $2,282.40