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Viefleur [7K]
3 years ago
6

This problem is taken from the delightful book "Problems for Mathematicians, Young and Old" by Paul R. Halmos. Suppose that 681

tennis players want to play an elimination tournament. That means: they pair up, at random, for each round; if the number of players before the round begins is odd, one of them, chosen at random, sits out that round. The winners of each round, and the odd one who sat it out (if there was an odd one), play in the next round, till, finally, there is only one winner, the champion. What is the total number of matches to be played together, in all the rounds of the tournament
Mathematics
1 answer:
aleksley [76]3 years ago
4 0

Answer:

680 games

Step-by-step explanation:

Suppose that 681 tennis players want to play an elimination tournament.

1st round:

One of 681 players, chosen at random, sits out that round and 680 players play. There will be 340 winners plus one player which sits - 341 players for the next round and 340 games

2nd round:

There will be 170 winners plus one player which sits - 171 players for the next round and 170 games

3rd round:

There will be 85 winners plus one player which sits - 86 players for the next round and 85 games

4th round:

There will be 43 winners - 43 players for the next round and 43 games

5th round:

There will be 21 winners plus one player which sits - 22 players for the next round and 21 games

6th round:

There will be 11 winners  - 11 players for the next round and 11 games

7th round:

There will be 5 winners plus one player which sits - 6 players for the next round and 5 games

8th round:

There will be 3 winners  - 3 players for the next round and 3 games

9th round:

There will be 1 winner plus one player which sits - 2 players for the next round and 1 game

10th round - final:

1 champion and 1 game.

In total,

340 + 170 + 85 + 43 + 21 + 11 + 5 + 3 + 1 + 1 = 680 games

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Which statement correctly explains whether or not the values in the table represent a function of x
Lemur [1.5K]

Answer:

Step-by-step explanation:

A relation is said to be a function when each input value (x-value) show a distinct output value (y-value).

From the table attached,

For input value x = 0, there are multiple values of output values (y = -2, -1, 0, 1, -2).

Therefore, data given in the table doesn't represent a function.

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3 years ago
Solve for r. p = 4r 3t
lara31 [8.8K]
P = 4r + 3t
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3 years ago
What notation is used to represent the distance from point A to B
borishaifa [10]

Answer:

Absolute Value

Step-by-step explanation:

Absolute Value is very useful in finding the distance between two points on the number line. The distance between any two points a and b in the number line is

|a-b| or |b-a|.

7 0
2 years ago
Samantha is using a 2-liter pitcher to serve lemonade to 10 of her friends. How many times will she need to fill the pitcher in
Mice21 [21]

Answer: 2 times

Step-by-step explanation:

Hi, since 1000 milliliters = 1 liter

We have to convert the 2 liters into milliliters:

2 x 1000 = 2000 milliliters, so she is using a 2,000 milliliters pincher.

Now we have to multiply the number of friends (10) by the amount of lemonade she serves to each one (400 ml)

400 x 10 = 4,000 milliliters

Finally, we have to divide the amount of lemonade she needs to serve (4,000 ml) by the capacity of the pitcher (2,000), to obtain the number of times that she will need to fill the pitcher.

4,000/2,000=2 times

Feel free to ask for more if needed or if you did not understand something.

4 0
3 years ago
Read 2 more answers
Use the Newton-Raphson method to find the root of the equation f(x) = In(3x) + 5x2, using an initial guess of x = 0.5 and a stop
xxMikexx [17]

Answer with explanation:

The equation which we have to solve by Newton-Raphson Method is,

 f(x)=log (3 x) +5 x²

f'(x)=\frac{1}{3x}+10 x

Initial Guess =0.5

Formula to find Iteration by Newton-Raphson method

  x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\\\\x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}\\\\ x_{1}=0.5-\frac{\log(1.5)+1.25}{\frac{1}{1.5}+10 \times 0.5}\\\\x_{1}=0.5- \frac{0.1760+1.25}{0.67+5}\\\\x_{1}=0.5-\frac{1.426}{5.67}\\\\x_{1}=0.5-0.25149\\\\x_{1}=0.248

x_{2}=0.248-\frac{\log(0.744)+0.30752}{\frac{1}{0.744}+10 \times 0.248}\\\\x_{2}=0.248- \frac{-0.128+0.30752}{1.35+2.48}\\\\x_{2}=0.248-\frac{0.17952}{3.83}\\\\x_{2}=0.248-0.0468\\\\x_{2}=0.2012

x_{3}=0.2012-\frac{\log(0.6036)+0.2024072}{\frac{1}{0.6036}+10 \times 0.2012}\\\\x_{3}=0.2012- \frac{-0.2192+0.2025}{1.6567+2.012}\\\\x_{3}=0.2012-\frac{-0.0167}{3.6687}\\\\x_{3}=0.2012+0.0045\\\\x_{3}=0.2057

x_{4}=0.2057-\frac{\log(0.6171)+0.21156}{\frac{1}{0.6171}+10 \times 0.2057}\\\\x_{4}=0.2057- \frac{-0.2096+0.21156}{1.6204+2.057}\\\\x_{4}=0.2057-\frac{0.0019}{3.6774}\\\\x_{4}=0.2057-0.0005\\\\x_{4}=0.2052

So, root of the equation =0.205 (Approx)

Approximate relative error

                =\frac{\text{Actual value}}{\text{Given Value}}\\\\=\frac{0.205}{0.5}\\\\=0.41

 Approximate relative error in terms of Percentage

   =0.41 × 100

   = 41 %

7 0
2 years ago
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