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sukhopar [10]
2 years ago
14

En una olimpiada matemática, cada estudiante recibe un punto por cada pregunta acertada. En la siguiente tabla de frecuencias se

muestra la clasificación de la cantidad de estudiantes de acuerdo con su puntaje obtenido. ¿Cuál es la probabilidad de seleccionar aleatoriamente un participante que haya obtenido entre 71 y 85 puntos?
Mathematics
1 answer:
Leviafan [203]2 years ago
4 0

La tabla relacionada con la pregunta se puede encontrar en la imagen adjunta a continuación:

Responder:

70%

Explicación paso a paso:

probabilidad de seleccionar aleatoriamente a un participante que haya obtenido entre 71 y 85 puntos:

Probabilidad = resultado requerido / Total de resultados posibles

Resultados posibles totales = Sumando la frecuencia para obtener el número total de estudiantes = (9 + 12 + 12 + 18 + 9) = 60 estudiantes

Resultado requerido:

Clase :

(71 - 75) = frecuencia = 12

(76 - 80) = frecuencia = 12

(81 - 85) = frecuencia = 18

Total = (12 + 12 + 18) = 42 = resultado requerido

Por tanto, P = 42/60 = 0,7 = (0,7 * 100%) = 70%

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X+y+z=12<br> 6x-2y+z=16<br> 3x+4y+2z=28<br> What does x, y, and z equal?
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Answer:

x = 20/13 , y = 16/13 , z = 120/13

Step-by-step explanation:

Solve the following system:

{x + y + z = 12 | (equation 1)

6 x - 2 y + z = 16 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Swap equation 1 with equation 2:

{6 x - 2 y + z = 16 | (equation 1)

x + y + z = 12 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Subtract 1/6 × (equation 1) from equation 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+(4 y)/3 + (5 z)/6 = 28/3 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Multiply equation 2 by 6:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Subtract 1/2 × (equation 1) from equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

0 x+5 y + (3 z)/2 = 20 | (equation 3)

Multiply equation 3 by 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

0 x+10 y + 3 z = 40 | (equation 3)

Swap equation 2 with equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+8 y + 5 z = 56 | (equation 3)

Subtract 4/5 × (equation 2) from equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+(13 z)/5 = 24 | (equation 3)

Multiply equation 3 by 5:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+13 z = 120 | (equation 3)

Divide equation 3 by 13:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Subtract 3 × (equation 3) from equation 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y+0 z = 160/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Divide equation 2 by 10:

{6 x - 2 y + z = 16 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Add 2 × (equation 2) to equation 1:

{6 x + 0 y+z = 240/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Subtract equation 3 from equation 1:

{6 x+0 y+0 z = 120/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Divide equation 1 by 6:

{x+0 y+0 z = 20/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Collect results:

Answer:  {x = 20/13 , y = 16/13 , z = 120/13

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A rectangle and a square have the same area. The length of the rectangle is seventy feet more than two times its width. The leng
Zanzabum

Using the side of the square find the area:

Area = 30^2 = 900 square feet.

The rectangles area is the same, 900 square feet.

Let the width = X

The length would be 2X + 70

Area = length x width

X * 2x+ 70 = 900

This expands to 2x^2 * 70x = 900

Use the quadratic formula to solve for x:

-70 +/- sqrt(70^2-4*2(-900))/2*2

X = 10

Width = x = 10 feet

Length = 2x + 70 = 90 feet

8 0
2 years ago
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