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Anuta_ua [19.1K]
3 years ago
6

Los estudiantes de grado decimo deciden realizar un paseo de despedida de fin de año. Han cotizado los precios para viajar a Car

tagena y encuentran que los tiquetes de ida y regreso cuestan 420000 y que en promedio cada día de estadía la cuesta 120000 por persona.
La función que representa el costo total por estudiante (y) en función de los días de estadía en Cartagena (x) es:
A. Y=420000 x +120000
B. Y=120000 x + 420000
C. Y=420000 x - 120000
D. Y=120000 x - 420000
Mathematics
1 answer:
nalin [4]3 years ago
3 0
The correct answer is B) y = 120000x + 420000.

We multiply 120,000 by x since that is the cost per person per day, and x is the number of days.  We add 420,000 because that is the one time cost of the round trip airfare.
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A ball is drawn at random from a box containing 12 red,18
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<u>Answer:</u>

<u>For a:</u> The probability of getting a red or blue ball is 0.48

<u>For b:</u> The probability of getting a white, blue or orange ball is 0.81

<u>For c:</u> The probability of getting neither white or orange ball is 0.48

<u>Step-by-step explanation:</u>

Probability is defined as the extent to which an event is likely to occurs. It is measured by the ratio of the favorable outcomes to the total number of possible outcomes.

\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Total number of favorable outcomes}}     .......(1)

We are given:

Number of red balls in a box = 12

Number of white balls in a box = 18

Number of blue balls in a box = 19

Number of orange balls in a box = 15

Total balls in a box = [12 + 18 + 19 + 15] = 64

  • <u>For a:</u>

Number of favorable outcomes (ball must be red or blue) = [12 + 19] = 31

Total number of outcomes = 64

Putting values in equation 1, we get:

\text{Probability of getting a red or blue ball}=\frac{31}{64}=0.48

  • <u>For b:</u>

Number of favorable outcomes (ball must be white or blue or orange) = [18 + 19 + 15] = 52

Total number of outcomes = 64

Putting values in equation 1, we get:

\text{Probability of getting a white or blue or orange ball}=\frac{52}{64}=0.81

  • <u>For c:</u>

Number of favorable outcomes (ball must be white or orange) = [18 + 15] = 33

Total number of outcomes = 64

Putting values in equation 1, we get:

\text{Probability of getting a white or orange ball}=\frac{33}{64}=0.52

Probability of getting a ball which is neither white or orange = [1 - (Probability of getting a white or orange ball)] = [1 - 0.52] = 0.48

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