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Home » Therefore, the number of permutations satisfying both conditions is: - Abbey Badges

Therefore, the number of permutations satisfying both conditions is: - Abbey Badges

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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\binom{6}{2} (0.20)^2 (0.80)^4 = 15 \cdot 0.04 \cdot 0.4096 = 0.24576
The probability is $\boxed{0.2458}$.Question: What is the remainder wh...
Solution: The first 5 terms are $ a_1 = 5 $, $ a_2 = 8 $, $ a_3 = 11 $...
Question: What is the least common multiple of the number of eggs laid...
Solution: Factorize each number: $ 12 = 2^2 \times 3 $, $ 18 = 2 \time...
\boxed{180}
Question: What two-digit number is one more than a multiple of 9 and a...
Solution: Let $ x \equiv 1 \pmod{9} $ and $ x \equiv 1 \pmod{11} $. Th...
\boxed{1} (Note: The question may require reevaluation for valid const...
Question: What is the largest integer that must divide the product of...

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