Let's analyze the equation \( m(p) = 20(2^p) \).
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Understanding the equation: Here, \( m \) (the amount of money) starts at 20 and doubles each time \( p \) (the number of people who pass the money to the next individual) increases. The factor \( 2^p \) indicates that as more people choose to pass the money, the amount increases exponentially.
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Range of \( m \): Since \( m = 20(2^p) \), and \( 2^p \) will always be greater than 1 for \( p \geq 0\), the minimum value of \( m \) occurs when \( p = 0\), which gives \( m = 20(2^0) = 20(1) = 20 \). As \( p \) increases, \( m \) becomes significantly larger due to the exponential growth of \( 2^p \). Therefore, the range of \( m \) includes values greater than 20: \( (m > 20) \) is true for \( p \geq 1 \).
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Behavior of \( m \): As \( p \) increases, \( m \) increases exponentially due to the \( 2^p \) term. This means \( m \) will increase quickly at first (for small \( p \)) and then continue to grow very quickly as \( p \) becomes larger, which contradicts the idea that it increases slowly at first then quickly.
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The domain of \( p \): The variable \( p \), which represents the number of people who pass on taking the money, starts at 0 (the initial choice to take the money or pass it). Therefore, the domain should include non-negative integers, \( p \geq 0 \), not starting from \( p \geq 20 \).
Based on the analysis above:
- The statement "The range (m-values) includes the values (m > 20)" is true.
- The statement "As p increases, m increases quickly at first and then slowly" is false.
- The statement "As p increases, m decreases slowly at first and then quickly" is false.
- The statement "The domain (p-values) include the values greater than or equal to 20 (p ≥ 20)" is false.
Conclusion: The correct and true statement about the equation is that the range (m-values) includes the values (m > 20).