Certainly! Let's go step-by-step:
1. Understand the problem:
- Mr. Rosewood has 15 gallons of paint.
- Each room needs [tex]\(1 \frac{2}{3}\)[/tex] gallons of paint.
- We need to determine how many classrooms he can paint with the available paint.
2. Convert the mixed number to an improper fraction:
- The mixed number [tex]\(1 \frac{2}{3}\)[/tex] can be converted to an improper fraction.
- Recall that [tex]\(1 \frac{2}{3} = 1 + \frac{2}{3} = \frac{3}{3} + \frac{2}{3} = \frac{5}{3}\)[/tex].
3. Set up the division problem:
- To find out how many classrooms can be painted, divide the total paint by the paint required per room.
- This translates to dividing 15 gallons by [tex]\(\frac{5}{3}\)[/tex] gallons.
4. Perform the division:
- Dividing by a fraction is the same as multiplying by its reciprocal.
- The reciprocal of [tex]\(\frac{5}{3}\)[/tex] is [tex]\(\frac{3}{5}\)[/tex].
[tex]\[
15 \div \frac{5}{3} = 15 \times \frac{3}{5}
\][/tex]
5. Simplify the multiplication:
[tex]\[
15 \times \frac{3}{5} = \frac{15 \times 3}{5} = \frac{45}{5} = 9
\][/tex]
So, Mr. Rosewood can paint 9 classrooms with the 15 gallons of paint he bought.
Therefore, the answer is (a) 9 classrooms.
