Select the correct answer.

Each month, Barry makes three transactions in his checking account:
- He deposits [tex]$\$700$[/tex] from his paycheck.
- He withdraws [tex]$\$150$[/tex] to buy gas for his car.
- He withdraws [tex]$\$400$[/tex] for other expenses.

If his account balance is [tex]$\$1,900$[/tex] at the end of the 1st month, which recursive equation models Barry's account balance at the end of month [tex]$m$[/tex]?

A. [tex]$f(1) = 1,900$[/tex]
[tex]$f(n) = f(n-1) - 150$[/tex], for [tex]$n \geq 2$[/tex]

B. [tex]$f(1) = 1,900$[/tex]
[tex]$f(n) = f(n-1) + 700$[/tex], for [tex]$n \geq 2$[/tex]

C. [tex]$f(1) = 1,900$[/tex]
[tex]$f(n) = f(n-1) + 150$[/tex], for [tex]$n \geq 2$[/tex]

D. [tex]$f(1) = 1,900$[/tex]
[tex]$f(n) = 150 \cdot f(n-1)$[/tex], for [tex]$n \geq 2$[/tex]

To solve the problem, we need to determine how Barry's account balance changes each month.

1. Barry deposits \[tex]$700 each month.
2. He withdraws \$[/tex]150 for gas.
3. He withdraws \[tex]$400 for other expenses.

The total amount withdrawn in a month is:
$[/tex][tex]$
150 + 400 = 550.
$[/tex][tex]$

The net change each month is the deposit minus the total withdrawals:
$[/tex][tex]$
700 - 550 = 150.
$[/tex][tex]$

We are given that the account balance at the end of the 1st month is \$[/tex]1,900. Therefore, the recursive equation that models the account balance is:
[tex]$$
f(1) = 1900,
$$[/tex]
and for every month [tex]$n \geq 2$[/tex]
[tex]$$
f(n) = f(n-1) + 150.
$$[/tex]

This corresponds to option C.