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vivado [14]
3 years ago
5

Un futbolista ha metido 2/9 del número de goles marcados por su equipo y otro la

Mathematics
1 answer:
storchak [24]3 years ago
7 0

Answer:

77 goles

Step-by-step explanation:

Un futbolista ha marcado 2/9 del número de goles marcados por su equipo.

Otro anotó una cuarta parte del resto.

El resto = 1 - 2/9

= 7/9

De ahí que otro futbolista anotó

= 1/4 de 7/9

= 7/36

Si los otros jugadores han marcado 45 goles

Tenemos que averiguar la fracción de goles que marcó el otro jugador

Deje que el número total de goles marcados por el equipo durante la temporada = 1

Por lo tanto:

1 - (2/9 + 7/36)

1 - (8 + 7/36)

1 - 15/36

1 - 5/12

= 7/12

¿Cuántos goles marcó el equipo a lo largo de la temporada?

El número total de goles que marcó ese equipo se calcula como:

7/12 × x = 45

Donde x = número total de goles

7x / 12 = 45

Cruz multiplicar

7x = 45 × 12

x = 45 × 12/7

x = 77,142857143

Aproximadamente = 77 goles

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Answer:

The value that represents the 90th percentile of scores is 678.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 550, \sigma = 100

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This is the value of X when Z has a pvalue of 0.9. So X when Z = 1.28.

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

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X - 550 = 100*1.28

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The value that represents the 90th percentile of scores is 678.

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3 years ago
If 3/x = 5/y , what is the value of y/x?<br><br><br> (15 POINTS AND FREE BRAINIEST ANSWER!!)
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3y/x=5  divide both sides by 3

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A college student is taking two courses. The probability she passes the first course is 0.73. The probability she passes the sec
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Answer:

b) No, it's not independent.

c) 0.02

d) 0.59

e) 0.57

f) 0.5616

Step-by-step explanation:

To answer this problem, a Venn diagram should be useful. The diagram with the information of Event 1 and Event 2 is shown below (I already added the information for the intersection but we're going to see how to get that information in the b) part of the problem)

Let's call A the event that she passes the first course, then P(A)=.73

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Then P(A∪B) is the probability that she passes the first or the second course (at least one of them) is the given probability. P(A∪B)=.98

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Two events are independent when P(A∩B) = P(A) * P(B)

So far, we don't know P(A∩B), but we do know that for all events, the next formula is true:

P(A∪B) = P(A) + P(B) - P(A∩B)

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c) The probability she does not pass either course, is 1 - the probability that she passes either one of the courses (P(A∪B) = .98)

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d) The probability she doesn't pass both courses is 1 - the probability that she passes both of the courses P(A∩B)

1 - P(A∩B) = 1 -.41 = .59

e) The probability she passes exactly one course would be the probability that she passes either course minus the probability that she passes both courses.

P(A∪B) - P(A∩B) = .98 - .41 = .57

f) Given that she passes the first course, the probability she passes the second would be a conditional probability P(B|A)

P(B|A) = P(A∩B) / P(A)

P(B|A) = .41 / .73 = .5616

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