Domain and range of y=x^2+2x+5

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:

<u>Trinomio Cuadrado Perfecto</u>

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

$(a-b)^2=a^2-2ab+b^2$

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

* El primer término es un cuadrado perfecto

* El último término es un cuadrado perfecto

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

Tenemos la expresión:

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

Calculamos el valor de a como la raiz cuadrada del primer término del trinomio:

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

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

Calculamos el valor de a como la raiz cuadrada del primer término del trinomio:

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

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

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:

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

5 0

3 years ago

Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe him. To persuad

laila [671]

Answer:

$17-2.262\frac{1.9}{\sqrt{10}}=15.641$

$17+2.262\frac{1.9}{\sqrt{10}}=18.359$

So on this case the 95% confidence interval would be given by (15.641;18.359)

And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=17$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.9 represent the sample standard deviation

n=10 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that $t_{\alpha/2}=2.262$