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3 years ago

10

Let us take as a given that x is normally distributed with a mean of 8.5 and a standard deviation of 2, what is p(x ≤ 6)? note:

round your answer, if necessary, to two places after the decimal. please express your answer with two places after the decimal.

1 answer:

n200080 [17]3 years ago

6 0

/ncvmbae,vnk.sdb,mfð v>Kjdbg,jrnvlnkjfnvkj,sbdkjvnkrjlhbvilabfns

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KengaRu [80]

Answer:

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Al factorizar el trinomio cuadrado perfecto, obtenemos el siguiente resultado: (que no se como resolver) xd algun pro que sepa r

PolarNik [594]

Answer:

$\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8=\left(\frac{10}{9}m^4p^{6}q^{8}z-\frac{1}{7}mpz^4\right)^2$

Step-by-step explanation:

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El producto notable llamado cuadrado de un binomio se expresa como:

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Si se tiene un trinomio, es posible convertirlo en un cuadrado perfecto si cumple con las condiciones impuestas en la fórmula:

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* El último término es un cuadrado perfecto

* El segundo término es el doble del proudcto de los dos términos del binomio.

Tenemos la expresión:

$\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8$

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$\displaystyle a=\sqrt{\frac{100}{81}m^8p^{12}q^{16}z^2}$

$\displaystyle a=\frac{10}{9}m^4p^{6}q^{8}z$

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$\displaystyle b=\sqrt{\frac{1}{49}m^2p^2z^8$

$\displaystyle b=\frac{1}{7}mpz^4$

Nos cercioramos de que el término central es 2ab:

$\displaystyle 2ab=2\frac{10}{9}m^4p^{6}q^{8}z\frac{1}{7}mpz^4$

Operando:

$\displaystyle 2ab=\frac{20}{63}m^5p^7q^8z^5$

Una vez verificado, ahora podemos decir que:

$\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8=\left(\frac{10}{9}m^4p^{6}q^{8}z-\frac{1}{7}mpz^4\right)^2$

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