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riadik2000 [5.3K]
3 years ago
10

There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the s

helf contain at least one paperback and at least one hardback? A75 b120 c210 d246 e252
Mathematics
1 answer:
FinnZ [79.3K]3 years ago
7 0

The answer is d (246)


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Answer:

1/4 gallon

Step-by-step explanation:

the least number that has at least one x is 1/4

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2 years ago
PLSSSS HELP<br><br> Which statement is true about the ratios shown?
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3 0
2 years ago
Read 2 more answers
Jerry started doing sit-ups every day. The first day he did 18 sit-ups. Every day after that he did 4 more sit-ups than he had d
USPshnik [31]

Answer:

18 + (4*5) = 38


because 4 * 5 is 20 then 20 + 18 is 38, and that's your answer.



have a nice day!   :)


7 0
3 years ago
. Imagine a game of 3 players where exactly one player wins in the end and all players have equal chances of being the winner. T
lbvjy [14]

Answer:

5/9

Step-by-step explanation:

Number of players = 3

number of times game is repeated = 4

P( any person wins a game ) = 1/3

P ( any person does not win a game ) = 1 - 1/3 = 2/3

P ( any person wins no game in 4 attempts ) = ( 2/3 )^4 = 16/81

<em>Note : each player has equal chance of winning </em>

<u>Find the probability that there is at least one person who wins no games </u>

lets represent the probability of each player not wining a game with alphabet A

A1 = player 1 wins no game

A2 = player 2 wins no game

A3 = players 3 wins no game

Applying the inclusion-exclusion formula

<em>P( A1 ∪ A2 U A3 )</em><em> = P(A1 ) + P(A2) + P(A3) - P( A1 ∩ A2 ) - P( A2 ∩ A3 ) - P( A1            ∩ A3 )  + P( A1 ∩ A2 ∩ A3 ) </em>

where

P( A1 ∩ A2 ) = P( A1 wins all games )

P ( A1 wins all games in 4 attempts ) = ( 1/3 )^4 = 1/81

P( A1 ∩ A2 ∩ A3 ) = P ( no players wins any game in 4 attempts ) = 0

Hence

P( A1 ∪ A2 U A3 ) = 16/81 + 16/81 + 16/81 - 1/81 - 1/81 - 1/81 - 0 = 5/9

6 0
2 years ago
−​50​​/54​​−40%+1.65
irga5000 [103]
The answer is - 0.53
5 0
3 years ago
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