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Home » So solution: \( x < -3 \) or \( x > 4 \) - Aurero

So solution: \( x < -3 \) or \( x > 4 \) - Aurero

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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Calculamos el coeficiente binomial:
\binom{6}{4} = \frac{6!}{4!2!} = \frac{720}{24 \cdot 2} = 15
\left(\frac{1}{2}\right)^6 = \frac{1}{64}
Entonces:
P(X = 4) = 15 \cdot \frac{1}{64} = \frac{15}{64}
\boxed{\dfrac{15}{64}}
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P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}

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Now from (5): $ b = 2 - 3a $
Substitute into (6):
12a + 3(2 - 3a) = 10 \Rightarrow 12a + 6 - 9a = 10 \Rightarrow 3a = 4...
Then $ b = 2 - 3 \cdot \frac{4}{3} = 2 - 4 = -2 $
From (2): $ \frac{4}{3} - 2 + c = 3 \Rightarrow c = 3 + 2 - \frac{4}{3...
And $ d = 2 $
p(x) = \frac{4}{3}x^3 - 2x^2 + \frac{11}{3}x + 2
Thus, the remainder when $ p(x) $ is divided by $ x^4 - 1 $ is simply...
\boxed{\frac{4}{3}x^3 - 2x^2 + \frac{11}{3}x + 2}
Question: Let $ a $ and $ b $ be complex numbers such that $ a + b = 4...

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