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Home » P(T|F) = \fracT)P(T)P(F) = \frac0.150.19 \approx 0.7895 - Crosslake

P(T|F) = \fracT)P(T)P(F) = \frac0.150.19 \approx 0.7895 - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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\( n = \frac{-1 \pm 41}{2} \).
A rectangular garden has a length 5 meters more than twice its width....
Let width be \( w \). Then length is \( 2w + 5 \).
Area: \( w(2w + 5) = 150 \).
\( 2w^2 + 5w - 150 = 0 \).
Using the quadratic formula, \( w = \frac{-5 \pm \sqrt{25 + 1200}}{4}...
\( w = \frac{-5 \pm 35}{4} \).
\( w = 7.5 \) (since width must be positive).
Length = \( 2 \times 7.5 + 5 = 20 \).
Dimensions are 7.5 m by 20 m.

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