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Sliva [168]
2 years ago
10

A chess tournament was being held. This single-elimination tournament (in which paired competitors

Mathematics
1 answer:
Sergeeva-Olga [200]2 years ago
5 0

Answer:

A total of 15 matches were played.

Step-by-step explanation:

To determine the number of matches that were played in the chess tournament, knowing that there are 16 participants who face each other in a tournament of direct elimination, the following logical reasoning must be carried out:

In the first round there are 16 players, so being matches between 2 players, there are 8 matches (16/2). In the second round, applying the same reasoning and taking into account that there are only 8 players left, since the others have been eliminated, there are 4 matches (8/2). The same in the semifinals, where there are 2 games (4/2), and finally, the final to a single game.

Therefore, since 8 + 4 + 2 + 1 equals 15, the number of matches played in the tournament is 15.

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The least common multiple is : 18

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The slope of the line is
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Find the sum of all two digit whole numbers which are divisible by 11?​
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Given two angles that measure 50 degrees and 80 degrees and side that measures 4 feet, how many triangles, if any, can be constr
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3 0
2 years ago
A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex
Olenka [21]

Answer:

a

 The  90% confidence interval that  estimate the true proportion of students who receive financial aid is

     0.533  <  p <  0.64

b

   n = 1789

Step-by-step explanation:

Considering question a

From the question we are told that

      The sample size is  n = 200

      The number of student that receives financial aid is k = 118

Generally the sample proportion is  

      \^ p = \frac{k}{n}

=>   \^ p = \frac{118}{200}

=>   \^ p = 0.59

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of \frac{\alpha }{2}  is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }

 =>E =  1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }

=>  E = 0.057

Generally 90% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

  =>  0.533  <  p <  0.64  

Considering question b

From the question we are told that

    The margin of error  is  E = 0.03

From the question we are told the confidence level is  99% , hence the level of significance is    

      \alpha = (100 - 99 ) \%

=>   \alpha = 0.01

Generally from the normal distribution table the critical value  of   is  

   Z_{\frac{\alpha }{2} } = 2.58

Generally the sample size is mathematically represented as      

        [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=>      n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )

=>      n = 1789

8 0
2 years ago
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