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\textTime_2 = \frac12080 = 1.5 \text hours - Network4

📅 May 18, 2026 👤 scraface

May 18, 2026

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\( P_0 = 1200 \), \( P(10) = 7680 \)
\( 7680 = 1200 e^{10k} \Rightarrow e^{10k} = \frac{7680}{1200} = 6.4 \...
\( 10k = \ln(6.4) \approx 1.856 \Rightarrow k \approx 0.1856 \)
\( P(20) = 1200 e^{0.1856 \times 20} = 1200 e^{3.712} \)
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