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Anuta_ua [19.1K]
3 years ago
14

12 people are standing in line to buy movie tickets. The theater is running a promotion by giving away 2 concession stand gift c

ards to the people standing in line.
What is the probability that the 2nd and 3rd people in line will get chosen for the gift cards?
Mathematics
2 answers:
Olin [163]3 years ago
5 0
Can you clarify, please. do you mean every 2nd and 3rd person?

Anvisha [2.4K]3 years ago
3 0
1 in 12 chance of second person getting given a ticket
Then a 1 in 11 chance of the third person being given one.
1/12*1/11 = 1/132 chance. However, if it doesn't have to be in order, then there's a 1/12 chance of the third person getting a ticket, then a 1/11 chance of the second person getting one, so another 1/132 chance. Add them together for 2/132, or 1/66 chance
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Expand (2x+2)^6<br> How would you find the answer using the binomial theorem?
Yanka [14]

Answer:

Step-by-step explanation:

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\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\

(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+  \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.

8 0
1 year ago
Use the following graph of the function f(x) = −3x4 − x3 + 3x2 + x + 3 to answer this question: graph of negative 3 x to the fou
Serggg [28]

Answer:

The average rate of change from x=0 to x=1 for f(x) is 0.

Step-by-step explanation:

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Hence, the average rate of change from x=0 to x=1 for f(x) is 0.

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Answer:

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