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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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Answer:

Part A: The variables are the amount charged per hour dog walking and the amount charged per hour babysitting.

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Step-by-step explanation:

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Every day, Jorge buys a lottery ticket. Each ticket has a probability of of winning a prize. After six days, what is the probabi
Romashka [77]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Every day, Jorge buys a lottery ticket. Each ticket has a 0.16 probability of winning a prize. After six days, what is the probability that Jorge has won at least one prize? Round your answer to four decimal places.

Answer:

The probability that Jorge has won at least one prize after six days is

P(at least 1 win) = 0.6487

Step-by-step explanation:

Every day, Jorge buys a lottery ticket which has a 0.16 chance of winning a prize.

We want to find out the probability that Jorge has won at least one prize after six days.

P(at least 1 win) = 1 - P(not winning for 6 days)

We know that the probability of winning is 0.16 then the probability of not winning is

P(not winning) = 1 - 0.16 = 0.84

For 6 days,

P(not winning for 6 days) = 0.84×0.84×0.84×0.84×0.84×0.84

P(not winning for 6 days) = 0.84⁶

P(not winning for 6 days) = 0.3513

Finally,

P(at least 1 win) = 1 - P(not winning for 6 days)

P(at least 1 win) = 1 - 0.3513

P(at least 1 win) = 0.6487

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2)\\(-2)(4)=-8\\(2)(-4)+-8\\-16:2=-8\\-16:(-2)=8\neq-8\\\\\text{All of the following expressions have the same value except}\ -16:-2.


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(-2)^4=(-2)(-2)(-2)(-2)=16\\-2^4=-(2)(2)(2)(2)=-16\neq16\\(-2)(-2)(-2)(-2)=16\\2^4=(2)(2)(2)(2)=16\\\\\text{All of the following are equivalent except}\ -2^4.

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