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Strike441
17 days ago
14

The following six students – Andy, Betty, Chris, Frank, Mark and Mary are standing in line outside of a classroom: Chris, the la

st one in line, is right behind Andy; Mark is between Mary and Betty; Frank, who is one ahead of Betty, is the first one in line. Where is Mary in the line? Second Third Fourth Fifth
Mathematics
1 answer:
Svet_ta [4.3K]17 days ago
6 0

Response:

(C) Fourth

Detailed reasoning:

We will create a list based on the given constraints.

Chris is last in the order, positioned right after Andy.

Frank holds the first position in line and is one spot ahead of Betty.

Mark stands in between Betty and Mary.

  1. Frank
  2. Betty
  3. Mark
  4. Mary
  5. Andy
  6. Chris
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Answer:

160/1001, 175/1001

Step-by-step explanation:

i) We calculate:

₈C₁ methods to select 1 new camera from a selection of 8

₆C₃ methods to select 3 refurbished cameras from a selection of 8

₁₄C₄ methods to select 4 cameras from the total of 14 cameras

The probability formula is:

P = ₈C₁ ₆C₃ / ₁₄C₄

P = 8×20 / 1001

P = 160 / 1001

P ≈ 0.160

ii) For at most one new camera, it means we want either one new camera or none at all. We've calculated the probability of selecting one new camera already. The probability of not selecting any new camera is equivalent to selecting 4 refurbished cameras:

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P = 15 / 1001

Therefore, the combined probability is:

P = 160/1001 + 15/1001

P = 175/1001

P ≈ 0.175

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10 days ago
Solve y=3bx-7x for x
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Y = 3bx - 7x
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\frac{y}{3b-7} =x

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15 days ago
steven has 9 different shirts 5 different hats 4 different scarves. Steven thinks that if he picks just two of the three items o
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Answer:

Steven is mistaken.

Step-by-step explanation:

Steven has

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He selects only two out of the three types of clothing. The combinations can be calculated as

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In total, there are

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As a result, 101 Steven is not correct.

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Answer:

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