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Natalija [7]
2 years ago
9

Help:

D0" id="TexFormula1" title="1.\ &x^4-61x^2+900=0 \\ 2. \ & x^4-25x^2+144=0" alt="1.\ &x^4-61x^2+900=0 \\ 2. \ & x^4-25x^2+144=0" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
yawa3891 [41]2 years ago
7 0

\bf{1) \  x^4-61x^2+900=0 }

Applying the factorization method, for example:

      \bf{ 0=(x^2)^2-61(x^2)+900=(x^2-36)(x^2-25)}

               \bf{\\ &=(x+6)(x-6)(x+5)(x-5);    }

              \bf{ x_1=6 \quad x_2=-6 \quad \ \ \ x_3=5 \quad x_4=5  }

\bf{2) \  x^4-25x^2+144=0  }

Applying the factorization method:

          \bf{0&=(x^2)^2-25(x^2)+144=(x^2-16)(x^2-9) }

             \bf{ =(x-4)(x+4)(x-3)(x+3); }

            \bf{ x_1=4 \quad x_2=-4 \quad x_3=3 \quad x_4=-3  }

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Answer:

Intervalo de confianza

= (0.294, 0.386)

Step-by-step explanation:

La fórmula para el intervalo de confianza para la proporción se da como

p ± z × √p (1 - p) / n

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De donde de la pregunta anterior

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Por lo tanto,

Intervalo de confianza =

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Intervalo de confianza

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= 0.2935765404

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= 0.3864234596

Hence:Intervalo de confianza

= (0.294, 0.386)

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