Answer:
A. $762.94
B. $896.81
C. $133.87
Step-by-step explanation:
Given function modelling Lisa's saving account:
[tex]\boxed{f(x)=250(1.25)^x}[/tex]
where x is the number of years.
Given table modelling Xavier's savings account:
[tex]\begin{array}{|c|c|c|c|c|}\cline{1-5} x & 0 & 1 & 2 & 3\\\cline{1-5} g(x) & 200 & 270 & 364.5 & 492.08\\\cline{1-5}\end{array}[/tex]
Part A
To find the amount in Lisa's savings account after 5 years, substitute x=5 into the function:
[tex]\begin{aligned}\implies f(5)&=250(1.25)^5\\&=250(3.051757...)\\&=762.939453...\end{aligned}[/tex]
Therefore, the amount in Lisa's savings account after 5 years is $762.94 (nearest cent).
Part B
First, create an exponential function to model Xavier's savings account.
General form of an exponential function:
[tex]y=ab^x[/tex]
where:
- a is the initial value (y-intercept).
- b is the base (growth/decay factor) in decimal form.
From inspection of the given table, the initial value (a) is 200.
[tex]\implies g(x)=200b^x[/tex]
To find the value of b, substitute point (1, 270) into the function:
[tex]\begin{aligned}\implies g(1)=200b&=270\\b&=\dfrac{270}{200}\\b&=1.35\end{aligned}[/tex]
Therefore. the function that models Xavier's savings account is:
[tex]\boxed{g(x)=200(1.35)^x}[/tex]
To find the amount in Xavier's savings account after 5 years, substitute x=5 into the found function:
[tex]\begin{aligned}\implies g(5)&=200(1.35)^5\\&=200(4.4840334...)\\&=896.806687...\end{aligned}[/tex]
Therefore, the amount in Xavier's savings account after 5 years is $896.81 (nearest cent).
Part C
To find the positive difference in their accounts after 5 years, subtract Lisa's balance from Xavier's balance:
[tex]\implies 896.81-762.94 =133.87[/tex]
