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puteri [66]
1 year ago
14

Clayton places tiles that spell the word time in a bowl. He draws one tile from the bowl and then without replacing it, draws an

other tile. What is the sample space of the event representing the letters picked in the two draws? {T-I, T-M, T-E, I-M, I-E, M-E} {T-I, T-M, T-E, I-T, I-M, I-E, E-T, E-M, E-I, M-I, M-T, M-E} {T-I, T-T, T-E, I-T, I-I, I-E, E-T, E-M, E-E, M-T, M-M, M-E} {T-I, T-M, T-E, T-T, I-T, I-M, I-E, I-I, E-T, E-M, E-I, E-E, M-T, M-T, M-E, M-M}

Mathematics
2 answers:
scoundrel [369]1 year ago
7 0
The problem can be represented in tree diagram as shown below. The combinations of drawing the first and the second tile is

TI      IT   MT   ET
TM   IM   MI    EI
TE    IE    ME  EM

There are few combinations that actually give the same result in any order, for example, TI and IT are the same combinations. We can narrow down the combination to:

TI 
TM 
TE
MI 
IE
ME
ET

lawyer [7]1 year ago
6 0

Answer:

The sample space of events that represent the picking out of letters at two draw without replacing is:

     {T-I, T-M, T-E, I-T, I-M, I-E, E-T, E-M, E-I, M-I, M-T, M-E}

Step-by-step explanation:

Since, there are four choice of drawing a first tile:

             T     I     M     E

and now:

  • If first tile is drawn as T then the outcome corresponding to T on second draw are:

                        T-I    T-M     T-E

(T-T is not possible since there is no replacement of first tile )

  • Similarly if I is drawn on first draw then the outcome corresponding to I in second draw is:

                      I-T     I-M     I-E

(I-I is not possible since there is no replacement of first tile )

  • If M is drawn on first draw then the outcome corresponding to M in second draw is:

                     M-T     M-I    M-E

(M-M is not possible since there is no replacement of first tile )

  • Now if E is drawn on first draw then the outcome corresponding to E in second draw is:

                         E-T     E-I      E-M    

(E-E is not possible since there is no replacement of first tile )

Hence, total outcomes are:

      {T-I, T-M, T-E, I-T, I-M, I-E, E-T, E-M, E-I, M-I, M-T, M-E}

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Snezhnost [94]

Answer:

There will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

The needed function is y = 2 ^n

Step-by-step explanation:

Let us assume the piece of paper James decides to fold is a SQUARE.

Now, let us assume:

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Now, if the paper if folded ONCE ⇒  n = 1

Also, when the pap er is folded once, the parts obtained are TWO equal parts.

⇒  for n = 1 , y = 2       ..... (1)

Similarly, if the paper if folded TWICE  ⇒  n = 2

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Continuing the same way, if the paper is folded SEVEN times  ⇒  n = 7

So, y = 2^ n = 2^7 = 128

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Hence, there will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

And the needed function is y = 2 ^n

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