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larisa86 [58]
2 years ago
5

Two students from a group of eight boys and 12 girls are sent to represent the school in a parade. If the students are chosen at

random, what is the probability that the students chosen are not both girls? StartFraction 12 Over 190 EndFraction StartFraction 33 Over 95 EndFraction StartFraction 62 Over 95 EndFraction StartFraction 178 Over 190 EndFraction
Mathematics
2 answers:
Tresset [83]2 years ago
5 0

The answer is C

62/95

kari74 [83]2 years ago
3 0

Answer:

Step-by-step explanation:

* Lets explain how to find the probability of an event  

- The probability of an Event = Number of favorable outcomes ÷ Total

 number of possible outcomes

- P(A) = n(E) ÷ n(S) , where

# P(A) means finding the probability of an event A  

# n(E) means the number of favorable outcomes of an event

# n(S) means set of all possible outcomes of an event

- Probability of event not happened = 1 - P(A)

- P(A and B) = P(A) . P(B)

* Lets solve the problem

- There is a group of students

- There are 8 boys and 12 girls in the group

∴ There are 8 + 12 = 20 students in the group

- The students are sent to represent the school in a parade

- Two students are chosen at random

∴ P(S) = 20

- The students that chosen are not both girls

∴ The probability of not girls = 1 - P(girls)

∵ The were 20 students in the group

∵ The number of girls in the group was 12

∴ The probability of chosen a first girl = 12/20

∵ One girl was chosen, then the number of girls for the second

  choice is less by 1 and the total also less by 1

∴ The were 19 students in the group

∵ The number of girls in the group was 11

∴ The probability of chosen a second girl = 11/19

- The probability of both girls is P(1st girle) . P(2nd girl)

∴ The probability of both girls = (12/20) × (11/19) = 33/95

- To find the probability of both not girls is 1 - P(both girls)

∴ P(not both girls) = 1 - (33/95) = 62/95

* The probability that the students chosen are not both girls is 62/95

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The size of fish is very important to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in n
devlian [24]

Answer:

The length of the longest 15% of Atlantic cod in this area is 53.79cm, roundeed to 2 decimal places.

Step-by-step explanation:

Problems of normally distributed samples can be 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}

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.

In this problem, we have that:

A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm, so \mu = 49.9, \sigma = 3.74.

What is the length in cm of the longest 15% of Atlantic cod in this area?

We have to find the value of X for the value of Z that has a pvalue of 0.85.

Looking at the zscore table, we have that Z = 1.04 has a pvalue of 0.8508. So, we have to find the value of X when Z = 1.04, \mu = 49.9, \sigma = 3.74.

So

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

1.04 = \frac{X - 49.9}{3.74}

X - 49.9 = 3.8896

X = 53.7896

The length of the longest 15% of Atlantic cod in this area is 53.79cm, roundeed to 2 decimal places.

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2) Which of the following describes the number of subscribers represented by the tables?
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How would u find 4 1\5 divided by 2 1\3
Over [174]
Remmber
(a/b)/(c/d)=(a/b)(d/c)=(ad)/(bc)
conver to improper

4 and 1/5=20/5+1/5=21/5
2 and 1/3=6/3+1/3=7/3

(21/5)/(7/3)=(21/5)(3/7)=63/35=9/5=1 and 4/5
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