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Home » Antwort: A $200e^{-0,5}$ - Aurero

Antwort: A $200e^{-0,5}$ - Aurero

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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Top 10 cm has 450 microfossils.
Density halves every 20 cm, so the decay has a half-life of 20 cm.
From 10–30 cm: depth interval = 20 cm → density halved → expected coun...
From 30–50 cm: another halving → 225 / 2 = 112.5
But we want 20–40 cm: this is 20 to 40, so 10–20 cm and 20–30 cm overl...
Better: define depth = d (cm from surface)
At d = 10, density = d₀ · (1/2)^(d/20)
Let N(d) = k × (1/2)^(d/20)
At d = 10, N(10) = 450 = k × (1/2)^(10/20) = k × (1/2)^(0.5) = k / √2
So k = 450 × √2 ≈ 450 × 1.4142 ≈ 636.39

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