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Elis [28]
2 years ago
13

On a sheet of paper, you have 100 statements written down. the first says, "at most 0 of these 100 statements are true." the sec

ond says, "at most 1 of these 100 statements are true." ... the nth says, "at most (n-1) of these 100 statements are true. ... the 100th says, "at most 99 of these statements are true." how many of the statements are true?
Mathematics
1 answer:
Yuki888 [10]2 years ago
4 0
I would go with the second statement is true because if all of the other ones mentioned in the problem had something that would not work either the one before or after would make the statement false excepted for statement two
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The positive trend among the homework scores and the test scores would reveal that if a student does well in homework, he is likely to do well in exams. This can be explained in the sense that the homework will serve as the practice for the student before finally taking the exam.

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Jessica has a bag with 2 mint sticks, 4 jelly treats, and 14 fruit tart chews. If she eats one piece every 9 minutes, what is th
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What are the coordinates of the image of vertex R -3,4 after a reflection across the y-axis? (–4, 3) (4, –3) (–3, –4) (3, 4)
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A way that landowners took advantage of sharecroppers was by:
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Rework problem 9 from section 3.2 of your text, involving independent and disjoint events. For this problem, assume that Pr[A∪B]
tatuchka [14]

Two events are said to be  Disjoint or Mutually Exclusive if the two events can not  happen at the same time.For example when we throw a die getting an even number is disjoint to getting an odd number.

I.e Probability(A∩B)=0

Let me explain this concept through venn diagram.

Pr[A∪B]=0.7, Pr[A]=0.25

Since events are Disjoint

Pr[A∩B]=0

Pr[A∪B]=Pr[A] + Pr[B]

0.7=0.25 +Pr[B]

0.7-0.25=Pr[B]

⇒Pr[B]=0.45=45/100=9/20

Now events are said to be independent if Pr[A and B]=Pr[A] ×Pr[B]

Events are said to be independent if occurrence of one is not affected by occurrence of other.For example getting multiple of 2 as one event and getting multiple of 3 as second event when we throw a die.

Pr[A∪B]=0.7, Pr[A]=0.25

Pr[A∪B]= Pr[A]+ Pr[B]-Pr[A∩B]

But Pr[A∩B]= Pr[A] ×Pr[B]

⇒Pr[A∪B]= Pr[A]+ Pr[B]- Pr[A] ×Pr[B]

⇒0.7=0.25+p-0.25×p

⇒0.7-0.25=p- 0.25 p

⇒0.45=0.75 p

⇒p= 0.45/0.75

⇒p =3/5


5 0
2 years ago
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