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Lena [83]
3 years ago
11

How many different tables of 4 can you make from 16 potential bridge players? How many different tables if 4 players insist on p

laying together?
Mathematics
2 answers:
Nadusha1986 [10]3 years ago
8 0

4 tables. 4*4 is 16, meaning that even if 4 players want to sit together it doesn't make a difference. With 16 players and 4 at a table we are going to need 4 tables.

xeze [42]3 years ago
7 0

Answer: Hence, a) 1820, b) 495.

Step-by-step explanation:

Since we have given that

Number of potential bridge players = 20

Number of tables = 4

We need to find the number of different tables of 4 from 16 potential bridge players.

So, Number of different ways would be

^{16}C_4\\\\=1820

If 4 players insist on playing together,

remaining players = 16-4 =12

so, Number of different ways would be

^{12}C_4=495

Hence, a) 1820, b) 495.

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9514 1404 393

Answer:

  1. asymptotes: x=-6, x=6, y=0; zero: x=0

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Step-by-step explanation:

1. The vertical asymptotes are where the denominator is zero. It factors as (x-6)(x+6), so will have zeros at x=-6 and x=6.

The horizontal asymptote is at the limit when x goes to infinity. Here, the ratio of highest-degree terms is 6/x, so the value goes to zero as x gets large.

The zero is where the numerator is zero, at x=0.

In summary:

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2. 60% are red, so 40% are blue. The difference in percentages is 60%-40% = 20%. The percentage of blue fish (40%) is twice the difference (=2×20%), so the number of blue fish is twice the difference in numbers.

  blue fish = 2×10 = 20 fish

<em><u>Check</u></em>

There are then 20+10=30 red fish, so 20+30=50 fish total. Reds are 30/50 = 60% of the total.

<u><em>Additional comment</em></u>

I find it often works well to work with ratios, which I can often do in my head. Other folks like to see equations. Here, we could write equations like ...

  r = 60%(r + b) . . . . reds are 60% of the total

  r - b = 10 . . . . . . . . 10 more reds than blues

We want the value of b, so we can eliminate r by substituting for it:

  r = 10 + b . . . . from the second equation above

  (10 +b) = 0.60((10 +b) + b) . . . . substitute for r

  10 +b = 6 + 1.2b . . . simplify

  4 = 0.2b . . . . subtract b+6

  20 = b . . . . . . multiply by 5 (there are 20 blue fish)

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