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Len
9 days ago
11

Two of the steps in the derivation of the quadratic formula are shown below. Step 6: StartFraction b squared minus 4 a c Over 4

a squared EndFraction = (x + StartFraction b Over 2 a EndFraction) squared Step 7: StartFraction plus or minus StartRoot b squared minus 4 a c EndRoot Over 1 a EndFraction = x + StartFraction b Over 2 a EndFraction Which operation is performed in the derivation of the quadratic formula moving from Step 6 to Step 7? subtracting StartFraction b Over 2 a EndFraction from both sides of the equation squaring both sides of the equation taking the square root of both sides of the equation taking the square root of the discriminant

Mathematics
2 answers:
babunello [3.6K]9 days ago
6 0

Explanation:

Step-by-step clarification:

Referring to step 6

(b² — 4ac) / 4a² = (x + b/2a)²

The mistake in the question is that it should be (x + b/2a)²

According to step 7

±√(b² —4ac) /2a = x + b/2a

The error in the question is that it should be divided by 2a, not 1a.

1. The transition from step 6 to step 7 involves taking the square roots of both sides

(b² — 4ac) / 4a² = (x + b/2a)²

Taking the square of both sides

√(b²—4ac) / √4a² = √(x + b/2a)²

√(b²—4ac) / 2a = x + b/2a

This forms step 7 correctly.

Next, subtracting b/2a from both sides

√(b²—4ac) / 2a - b/2a= x + b/2a -b/2a

√(b²—4ac) / 2a — b/2a = x

(√(b²—4ac)  — b)/2a = x

x = [—b ± √(b²—4ac)] / 2a

This gives the desired formula.

The discriminant is D = b²—4ac.

Svet_ta [4.3K]9 days ago
5 0

Explanation:

c

Detailed reasoning:

Square roots of both sides are extracted from the equation

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Answer:

a) 1.88% de los corredores tienen frecuencias cardíacas superiores a 130.

b) 50% de los no corredores tienen frecuencias cardíacas superiores a 130.

Explicación paso a paso:

Los problemas relacionados con muestras distribuidas normalmente pueden resolverse utilizando la fórmula del puntaje z.

En un conjunto con media \mu y desviación estándar \sigma, el puntaje z asociado a una medida X se presenta a través de:

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(a) ¿Qué porcentaje de los corredores tienen frecuencias cardíacas superiores a 130?

En un estudio sobre ejercicio, un gran grupo de corredores masculinos camina en una caminadora durante seis minutos. Sus frecuencias cardíacas expresadas en latidos por minuto al final varían entre ellos conforme a la distribución N(104,12.5). Esto implica que \mu = 104, \sigma = 12.5.

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(b) ¿Qué porcentaje de los no corredores tienen frecuencias cardíacas superiores a 130?

Las frecuencias cardíacas para hombres no corredores después del mismo ejercicio mantienen la distribución N(130, 17). Esto implica que \mu = 130, \sigma = 17.

Este porcentaje es 1 menos el valor p cuando X = 130.

Z = \frac{X - \mu}{\sigma}

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Z = 0.00

Z = 0.00 tiene un valor p de 0.5000.

Esto significa que 1-0.50 = 0.50 = 50% de los no corredores tienen frecuencias cardíacas superiores a 130.

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