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frez [133]
3 years ago
11

Please help mee dont answer if you dunno tho

Mathematics
1 answer:
Murrr4er [49]3 years ago
4 0

A). Find the explicit expression for this sequence

Ans:  A(n)=3n-2

B. Find the recursive expression for this sequence

Ans: A(n)=A(n-1)+3

C. What is the 15th term?

Ans: 15th term is 43

Step-by-step explanation:

Arithmetic sequence is given by a,a+d,a+2d,a+3d.......a+(n-1)d

Where a is first term of arithmetic sequence

Let,

The coordinates of x axis represent the location of term, say n

The coordinates of y axis represent the value of pespective term

We can write

Arithmetic sequence as 1,4,7...........

Where a=1 and a+d=4

For value of d,

d=(a+d)-(a)=(4)-(1)=3

A. Find the explicit expression for this sequence

Ans:

Explicit expression is given by A(n)=a+(n-1)d

For given Arithmetic sequence as 1,4,7...........

Explicit expression will be

A(n)=a+(n-1)d

A(n)=1+(n-1)3

A(n)=1+3n-3

A(n)=3n-2

B. Find the recursive expression for this sequence

Ans:

Recursive expression is given by A(n)=A(n-1)+d

For given Arithmetic sequence as 1,4,7...........

Recursive expression will be

A(n)=A(n-1)+d

A(n)=A(n-1)+3

C. What is the 15th term?

Ans:

For given Arithmetic sequence as 1,4,7...........

Explicit expression is given by

A(n)=3n-2

For 15th term n=15

A(n)=3(15)-2=45-2=43

Thus, 15th term is 43

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Answer:

The probability that 3 lights are on after the third person exited the room is 39/128

Step-by-step explanation:

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Note that after turning off one switch one more switch will be available to be switched in and one less will be available to be switch off. The contrary happens when someone turns in a switch.

Lets calculate the probability for the first scenario. The probability for the first person to turn the switch in is 3/8, because there are 3 lights off. For the second person, there will be only 2 lights off, thus, the probability for him or her to turn the switch in is only 2/8, leaving only 1 light off and 7 on. The third person will have, as a consecuence, a probability of 7/8 to turn off one of the 7 switches. This gives us a probability of 3/8 * 2/8 * 7/8 = 21/256 for the first scenario.

For the second scenario we will have a probability of 3/8 for the first person, a probability of 6/8 for the second one (he has to turn a switch off this time), and a probability, again, of 3/8 for the third one, giving us a probability of 3/8*6/8*3/8 = 27/256 for the second scenario.

For the third scenario, the first person has to turn off the switch, and it has a probability of 5/8 of doing so. The second person will have 4 switches to turn on, so it has a probability of 4/8 = 1/2, and the third person will have one switch less, thus, a probability of 3/8 of turning a switch on. Therefore, the probability of the third scenario is 5/8*1/2*3/8 = 15/128 = 30/256

By summing all the three disjoint scenarios, the probability that six lights are on is 21/256+27/256+30/256 = 78/256 = 39/128.

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Step-by-step explanation:

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