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3 years ago

El cuadrado de un número disminuido en 9 equivale a 8 veces el exceso

Mathematics

1 answer:

olga_2 [115]3 years ago

7 0

I think this was already answered over here.

X : EL NUmerRO DESCONOCIDO

X*2 : X ELEVADO AL CUADRADO

X*2 -9 : EL NUMERO ELEVADOAL CUADRADO Y DISMINUIDO EN 9

Equivale es igual a :

X -2 : EL EXCESO DEL NUMERO SOBRES DOS (se deduce que X es mayor a 2)

8(X-2) : ocho veces el exceso del numero sobre 2

ECUACION

x*2 -9 = 8 (X-2)

X*2 -9=8X-16

X*2 - 8X -9+16=0

X*2 - 8X +7 =0

pOR EL METODO DE ASPA SIMPLE Y COMO ES CUADRATICA X SALE CON DOS VALORES

X= 7 ó X=1

LA RESPUESTA ES X=7 YA QUE DE LA CONDICION SABEMOS QUE X TENE QUE SER mayor que 2

RESPUESTA EL NUMERO ES 7

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3 years ago

In a survey of 447 registered voters, 157 of them wished to see Mayor Waffleskate lose her next election. The Waffleskate campai

Hitman42 [59]

<h2>Answer with explanation:</h2>

Let p be the true proportion of registered voters wish to see Mayor Waffleskate defeated.

As per given , we have

$H_0: p\leq0.27\\\\ H_a: p >0.27$

Sample size : n= 447

Number of of registered voters wish to see Mayor Waffleskate defeated = 157

I.e. sample proportion : $\hat{p}=\dfrac{157}{447}\approx0.3512$

Confidence interval for population proportion is given by :-

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where n= sample size

$\hat{p}$ = sample proportion

z* = critical z-value.

Critical z-value for 98% confidence interval is 2.33. (By z-table)

Then, the 98% confidence interval for the proportion of registered voters who wish to see Waffleskate defeated will be :

$0.3512\pm2.33\sqrt{\dfrac{0.3512(1-0.3512)}{447}}\\\\=0.3512\pm (2.33)(0.022577656)\\\\=0.3512\pm 0.05260593848\\\\=(0.3512-0.05260593848,\ 0.3512+0.05260593848)\\\\=(0.29859406152,\ 0.40380593848)\approx(0.299,\ 0.404)$

Since the 0.27 < 0.299 , it means 0.27 does not belong to the above confidence interval.

So , we reject the null hypothesis ( $H_0$ ).

So , <u>98% confidence interval does not support the claim.</u>

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