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Andrej [43]
3 years ago
9

Acertijo.

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
8 0

La edad de cada uno de los integrantes del acertijo es:

  • <u>Luisa = 3 años </u>
  • <u>Juan = 6 años </u>
  • <u>Carmen = 8 años</u>

<em>Cálculo por medio de ecuaciones y el </em><em>método de igualación</em>.

Haciendo uso de la información brindada en el ejercicio, se procede a formular variables con las cuales se calcularán las ecuaciones, por lo tanto:

  • <em>X = Edad de Luisa</em>
  • <em>Y = Edad de Juan</em>
  • <em>Z = Edad de Carmen</em>

Con el enunciado: <em>"Luisa es 3 años más joven que su hermano Juan"</em> se crea la primera ecuación:

  • 1. Y = X + 3

Con el enunciado: <em>"Su hermana Carmen es 2 años mayor que Juan"</em>, se crea la segunda ecuación:

  • 2. Z = Y + 2

Y con el enunciado: <em>"Juan tiene el doble de edad que Luisa"</em>, se crea la tercera ecuación:

  • 3. X = \frac{Y}{2}

Ahora, utilizando el método de igualación, se reemplaza la <em>tercera ecuación en la primera</em>:

  • Y = X + 3
  • Y = (\frac{Y}{2}) + 3

Y se procede a sumar:

  • Y = (\frac{Y}{2}) + 3
  • Y = \frac{Y+6}{2}

El dividendo se pasa al otro lado de la igualdad a multiplicar:

  • 2Y = Y + 6

Y la variable "Y" que está sumando, se pasa al otro lado de la igualdad a restar:

  • 2Y - Y = 6
  • <u>Y = 6</u>

Con el valor de Y (edad de Juan) <em>se calcula X con la tercera ecuación</em>:

  • X = \frac{Y}{2}
  • X = \frac{6}{2}
  • <u>X = 3</u>

Una vez calculado el valor de X (edad de Luisa), se procede a <em>calcular el valor de Z con la segunda ecuación</em>:

  • Z = Y + 2
  • Z = 6 + 2
  • <u>Z = 8</u>

Por lo tanto, se calcula que <u>la edad de Luisa es 3 años, la edad de Juan es 6 años y la edad de Carmen es 8 años</u>.

Más información sobre ecuaciones:

brainly.com/question/15415929?referrer=searchResults

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The probability that fewer than 4 of then use their smart phones in meetings  or classes is 0.00030243747.

Step-by-step explanation:

Fewer than 4 means ,less than 4 which is any number less than 4 and not including 4. This is 3, 2,1 and 0. So you find the probability that exactly 3 of then use their smart phone, exactly 2 of them use their smart phones, exactly 1 of them use his/her smart phone and none of them.

Given that 52% use smartphones in meetings and classes =0.52

Thus the remaining 48% do not use smartphones in meetings and classes=0.48

For C(14,3) you will have "14 chose 3", number of ways of choosing 3 out of 14'

=(0.52)³ *(0.48)⁹=0.00019018714

For C(14,2) you will have "14 chose 2", number of ways of choosing 2 out of 14

=(0.52)²*(0.48)¹²=0.00004044841

For C(14,1) you will have "14 chose 1", number of ways of choosing 1 out of 14

=(0.52)¹*(0.48)¹³=0.000037337

For  C(14,0) you will have "14 chose 0", number of ways of choosing 0 out of 14

=(0.52)⁰*(0.48)¹⁴=1*0.00003446492=0.00003446492

The probability that fewer than 4 of them use their smartphones in meetings or classes will be

0.00019018714 +0.00004044841+0.000037337+ 0.00003446492=0.00030243747

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Step-by-step explanation:

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<u>step 2:</u> After 4 hours, the number in the mile marker is 300

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