Question

1. 45% of the players on the soccer team got to practice early! if 5 students got to practice early, how many total players are on the team?

Explanation:

Step1: Define the variable

Let \( x \) be the total number of players on the team.

Step2: Set up the equation

We know that \( 45\% \) of \( x \) is equal to 5. Mathematically, this can be written as \( 0.45x = 5 \).

Step3: Solve for \( x \)

To find \( x \), we divide both sides of the equation by \( 0.45 \): \( x=\frac{5}{0.45}=\frac{5}{\frac{45}{100}} = 5\times\frac{100}{45}=\frac{100}{9}\approx11.11 \). But since the number of players must be a whole number, there might be a mistake in the problem statement or we assume it's a approximation. Wait, maybe I misread, 45%? Wait, no, maybe it's 40%? Wait, no, the problem says 45%. Wait, 5 divided by 0.45 is approximately 11.11. But maybe the problem has a typo, but following the math, \( x=\frac{5}{0.45}=\frac{100}{9}\approx11 \) (if we round) or maybe the percentage is 40%? Wait, no, the problem says 45%. Wait, maybe I made a mistake. Wait, 45% of x is 5. So \( x = 5\div0.45=\frac{5}{\frac{9}{20}}=5\times\frac{20}{9}=\frac{100}{9}\approx11.11 \). So the total number of players is approximately 11 (if we take the nearest whole number) or maybe the problem intended a different percentage, like 50%? But as per the given problem, it's 45%. So the calculation is \( x=\frac{5}{0.45}=\frac{100}{9}\approx11.11 \), so we can say approximately 11 or 12? Wait, no, maybe the problem has a mistake, but following the math, the exact value is \( \frac{100}{9}\approx11.11 \).

Wait, maybe I misread the percentage. Let me check again. The problem says 45% of the players got to practice early, and 5 students got early. So 0.45x = 5. So x = 5 / 0.45 = 100/9 ≈ 11.11. So the total number of players is approximately 11 (if we round down) or 12 (if we round up). But since we can't have a fraction of a player, maybe the problem has a typo, but mathematically, the answer is \( \frac{100}{9}\approx11.11 \).

Answer:

\(\boxed{\frac{100}{9}}\) (or approximately 11)