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Home » Stop Dying to Save: Here’s How 1 Million Will Change Your Life Forever - Get link 4share

Stop Dying to Save: Here’s How 1 Million Will Change Your Life Forever - Get link 4share

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 Let $ w = 1 - u $, $ dw = -du $, limits $ w = 0.2 $ to $ 0 $:

📌 \int_0^{0.2} w^5 dw = 6 \cdot \frac{w^6}{6} \Big|_0^{0.2} = (0.2)^6 = \frac{1}{15625}

📌 Total: $ \frac{1}{15625} + \frac{1}{15625} = \frac{2}{15625} $

📌 Thus, the probability that all 6 vehicles arrive within some 2-minute interval is:

📌 \boxed{\frac{2}{15625}}

📌 Question: A programmer is testing a neural network that generates binary strings of length 8, where each bit is independently 1 with probability $ \frac{1}{3} $ and 0 with probability $ \frac{2}{3} $. What is the probability that the string contains exactly three 1s, with no two 1s adjacent?

📌 Solution: We want the probability that a binary string of length 8 has exactly three 1s, no two of which are adjacent, with each 1 occurring independently with probability $ \frac{1}{3} $, and 0s with $ \frac{2}{3} $.

📌 First, compute the number of binary strings of length 8 with exactly 3 non-adjacent 1s.

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Solution: We use the well-known algebraic identity:
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca).
Given that $ a + b + c = 0 $, substitute into the identity:
a^3 + b^3 + c^3 - 3abc = 0 \cdot (a^2 + b^2 + c^2 - ab - bc - ca) = 0.
a^3 + b^3 + c^3 = 3abc.
This completes the proof.
oxed{a^3 + b^3 + c^3 = 3abc}
Question: Compute $ \sum_{k=1}^{50} rac{1}{k(k+2)} $.
Solution: We evaluate the sum:
\sum_{k=1}^{50} rac{1}{k(k+2)}.

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