### Solution to Problem 5:

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• ### Solution to Problem 4:
• First, compute \( g(3) = 2(3) - 5 = 6 - 5 = 1 \)
• Then \( f(g(3)) = f(1) = 3(1)^2 - 4(1) + 7 = 3 - 4 + 7 = 6 \)
• Let total students = \( x \)
• \( 0.65x = 1440 \Rightarrow x = \frac{1440}{0.65} \approx 2215.38 \)
• Since number of students must be whole, round to nearest integer: 2,215

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