Answer:
110 calendars to have a profit of atleast 1200.
Step-by-step explanation:
Initial expense = 600
Let the number of calendars that the photography club must sell in order to raise at least 1200$ be x.
Cost per calendar = 2.5
Cost for x calendars = 2.5 x
Total expenses = 2.5 x + 600
Income per calendar = 8
Income = 8 x
Minimum profit = 1200
Therefore the inequality is,
8x - 2.5 x + 600 ≥ 1200
Solving we get,
5.5x ≥ 600
x ≥ 109.09
x ≥ 110