Question 4 • Algebra 2 with Probability
Seven buses can carry at least 360 students only if at least 3 are large: with 3 large and 4 small, capacity is 3 × 60 + 4 × 45 = 360. That arrangement costs 3 × $300 + 4 × $200 = $1,700. Using fewer than seven buses cannot be cheaper while meeting the capacity requirement: six buses would need to all be large to carry 360 students, costing $1,800. So the minimum is choice D, $1,700.
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