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Maurinko [17]
3 years ago
12

Yats of fle har earh par O2 yr 03 ya 04yads 111

Mathematics
2 answers:
Vesna [10]3 years ago
8 0

Answer:

habibi astra madookoo a1 x 4x doodoo perosetro ekebodina daiki

Step-by-step explanation:

Marianna [84]3 years ago
6 0

Answer:huh?

Step-by-step explanation:

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Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that
aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

A\cap B={HH}

B\cap C={HH}

A\cap C={HH}

P(E)=\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

P(A)=\frac{2}{4}=\frac{1}{2}

Number of favorable cases to event B=2

Number of favorable cases to event C=2

P(B)=\frac{2}{4}=\frac{1}{2}

P(C)=\frac{2}{4}=\frac{1}{2}

If the two events A and B are independent then

P(A)\cdot P(B)=P(A\cap B)

P(A\cap)=\frac{1}{4}

P(B\cap C)=\frac{1}{4}

P(A\cap C)=\frac{1}{4}

P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}

P(B)\cdot P(C)=\frac{1}{4}

P(A)\cdot P(C)=\frac{1}{4}

P(A)\cdot P(B)=P(A\cap B)

Therefore, A and B are independent

P(B)\cdot P(C)=P(B\cap C)

Therefore, B and C are independent

P(A\cap C)=P(A)\cdot P(C)

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

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3 years ago
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Marysya12 [62]
It's impossible because there is no "N" point in rectangle JKLM.
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Please Help me! 20 points
malfutka [58]

Answer:

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Step-by-step explanation:

-(2 1/4) ➡ -9/4

-2/3 ➡ (again a reciprocal needed) -3/2 since we need the result to be positive both expression should remain negative so the answer is -9/4 ÷ -2/3

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Im thinking the answer is c
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