Question

at the end of each round of a basketball tournament, half of the teams are eliminated from the competition. if the tournament starts with 64 teams, which equation shows how many rounds, r, must be completed to have 2 teams left in the tournament?
64(\frac{1}{2})^{r}=2
2(2)^{r/2}=64
2(\frac{1}{2})^{r}=64
64(2)^{r}=\frac{1}{2}

Explanation:

Step1: Analyze the exponential decay model

The general formula for exponential decay is \(N = N_0(\frac{1}{2})^r\), where \(N_0\) is the initial amount, \(r\) is the number of time - periods (here, rounds), and \(N\) is the final amount.
Given \(N_0 = 64\) (initial number of teams) and \(N = 2\) (final number of teams).
Substitute these values into the formula: \(64(\frac{1}{2})^r=2\).

Step2: Check other options

• For \(2(2)^{r/2}=64\), if we simplify \(2(2)^{r/2}=2^{1 + r/2}\). Setting \(2^{1 + r/2}=64 = 2^6\), then \(1+\frac{r}{2}=6\), \(\frac{r}{2}=5\), \(r = 10\). But the model should be based on halving each round (\(y=a(\frac{1}{2})^x\) form).
• For \(2(\frac{1}{2})^r=64\), \(2(\frac{1}{2})^r=\frac{2}{2^r}\). \(\frac{2}{2^r}=64\), \(2^r=\frac{2}{64}=\frac{1}{32}\), \(r=- 5\) which is not meaningful for the number of rounds.
• For \(64(2)^r=\frac{1}{2}\), \(2^r=\frac{1}{128}\), \(r=-7\) which is not meaningful for the number of rounds.

Answer:

\(64(\frac{1}{2})^r = 2\) (the first option)