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julia-pushkina
12 days ago
7

Point P is in the interior of ∠OZQ. If m∠OZQ = 125 and m∠OZP = 62, what is m∠PZQ?

Mathematics
1 answer:
tester [3.9K]12 days ago
3 0

Response:

63°

Detailed explanation:

If point P lies within ∠OZQ, it holds that ∠OZQ equals the sum of ∠OZP and ∠PZQ. Given m∠OZQ = 125 and m∠OZP = 62, we can find m∠PZQ by substituting the known angle values into the aforementioned equation as illustrated;

∠OZQ = ∠OZP + ∠PZQ.

125 = 62 + ∠PZQ.

We subtract 62 from each side;

125 - 62 = 62 + ∠PZQ - 62

63 = ∠PZQ

Thus, the angle m∠PZQ measures 63°

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Find the radius of an aluminum cylinder that is 2.00 cm long and has a mass of 12.4 g.
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In a study of exercise, a large group of male runners walk on a treadmill for six minutes. Their heart rates in beats per minute
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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:

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

El puntaje z indica cuántas desviaciones estándar está la medida respecto a la media. Tras determinar el puntaje z, consultamos la tabla de puntajes z y localizamos el valor p relacionado con este puntaje. Este valor p representa la probabilidad de que el valor de la medida sea menor que X, es decir, el percentil de X. Al restar 1 del valor p, obtenemos la probabilidad de que el valor de la medida sea mayor que X.

(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.

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

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

Z = \frac{130 - 104}{12.5}

Z = 2.08

Z = 2.08 tiene un valor p de 0.9812.

Esto significa que 1-0.9812 = 0.0188 = 1.88% de los corredores tienen frecuencias cardíacas superiores a 130.

(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}

Z = \frac{130 - 130}{17}

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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5 days ago
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