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]n[/tex]?

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

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

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

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

To determine which recursive equation models Barry's account balance, let's break down his monthly transactions:

1. Deposits:
- Barry deposits [tex]$700 each month from his paycheck.

2. Withdrawals:
- He withdraws $[/tex]150 for gas.
- He also withdraws [tex]$400 for other expenses.

3. Net Monthly Change:
- The net effect on Barry's account each month can be calculated as:
\[
700 - 150 - 400 = 150
\]
- This means his account balance increases by $[/tex]150 each month.

Now, let's establish the recursive equation:

- At the end of the 1st month, Barry's account balance is [tex]$1,900. This is given as:
\[
f(1) = 1900
\]

- For subsequent months ($[/tex]n \geq 2[tex]$), the account balance can be found by adding the net monthly change ($[/tex]150) to the previous month's balance. Therefore, the recursive formula is:
[tex]\[
f(n) = f(n-1) + 150
\][/tex]

Comparing this with the provided options, this matches Option B:
- [tex]$f(1) = 1,900$[/tex]
- [tex]$f(n) = f(n-1) + 150$[/tex], for [tex]$n \geq 2$[/tex]

Therefore, the correct recursive equation is modeled by option B.