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Musya8 [376]
3 years ago
13

In a class of 200 students, 120 students take german, and 100 study french. if a student must study at least one of these langua

ges, what percent of the students study french but not german
Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
8 0
U=200
n(G)=120
n(F)=100
n(GnF)=x
U=n(G)+n(F)-n(GnF)
200=120+100-x
x=20
French but not German is 100-20=80
percentage = 80/200 *100%= 40%
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Step-by-step explanation:

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State police believe that 70% of the drivers traveling on a major interstate highway exceed the speed limit. They plan to set up
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Answer:

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Binomial probability distribution

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A professor has been teaching accounting for over 20 years. From her experience she knows that 75% of her students do homework r
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Answer:

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b) In order to two events can be considered as independent we need to have this condition:

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If we check the condition we see this:

P(A)*P(B)= 0.75*0.7=0.525 \neq P(A and B)

So since we don't have the condition satisfied the events "do the homework regularly" and "pass the course" are not independent.

Other way to check independence is with the following two conditions:

P(B|A)=P(B) or P(A|B)=P(A), and we see that P(B|A)\neq P(B)  for our case.

Step-by-step explanation:

Data given

Lets define the following events

A: Represent the students who do homework regularly

B: Represent the students who  pass the course

75% of her students do homework regularly

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For this case we have some probabilities given:

P(A) =0.75, P(B) =0.70 and P(A and B)=0.65

Solution to the problem

Part a

On this case we want to find this probability:

P(B|A) the conditional probability (Pass the course given that he does the homework regularly).

By defintion of conditional probability we know that:

P(B|A) =\frac{P(A and B)}{P(A)}

And now we can replace in the conditional formula like this:

P(B|A) =\frac{P(A and B)}{P(A)}=\frac{0.65}{0.75}=0.867

Part b

In order to two events can be considered as independent we need to have this condition:

P(A and B) = P(A)*P(B)

If we check the condition we see this:

P(A)*P(B)= 0.75*0.7=0.525 \neq P(A and B)

So since we don't have the condition satisfied the events "do the homework regularly" and "pass the course" are not independent.

Other way to check independence is with the following two conditions:

P(B|A)=P(B) or P(A|B)=P(A), and we see that P(B|A)\neq P(B)  for our case.

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In this problem, we have that:

\mu = 3, \sigma = 1

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6 0
3 years ago
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