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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?

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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3 years ago

If the average yield of cucumber acre is 800 kg, with a variance 1600 kg, and that the amount of the cucumber follows the normal

lorasvet [3.4K]

Answer:

a

The percentage is

$P(x_1 < X < x_2 ) = 51.1 \%$

b

The probability is $P(Z > 2.5 ) = 0.0062097$

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 800$

The variance is $var(x) = 1600 \ kg$

The range consider is $x_1 = 778 \ kg \ x_2 = 834 \ kg$

The value consider in second question is $x = 900 \ kg$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{var (x)}$

substituting value

$\sigma = \sqrt{1600}$

$\sigma = 40$

The percentage of a cucumber give the crop amount between 778 and 834 kg is mathematically represented as

$P(x_1 < X < x_2 ) = P( \frac{x_1 - \mu }{\sigma} < \frac{X - \mu }{ \sigma } < \frac{x_2 - \mu }{\sigma } )$

Generally $\frac{X - \mu }{ \sigma } = Z (standardized \ value \ of \ X)$

So

$P(x_1 < X < x_2 ) = P( \frac{778 - 800 }{40} < Z< \frac{834 - 800 }{40 } )$

$P(x_1 < X < x_2 ) = P(z_2 < 0.85) - P(z_1 < -0.55)$

From the z-table the value for $P(z_1 < 0.85) = 0.80234$

and $P(z_1 < -0.55) = 0.29116$

So

$P(x_1 < X < x_2 ) = 0.80234 - 0.29116$

$P(x_1 < X < x_2 ) = 0.51118$

The percentage is

$P(x_1 < X < x_2 ) = 51.1 \%$

The probability of cucumber give the crop exceed 900 kg is mathematically represented as

$P(X > x ) = P(\frac{X - \mu }{\sigma } > \frac{x - \mu }{\sigma } )$

substituting values

$P(X > x ) = P( \frac{X - \mu }{\sigma } >\frac{900 - 800 }{40 } )$

$P(X > x ) = P(Z >2.5 )$

From the z-table the value for $P(Z > 2.5 ) = 0.0062097$

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2 years ago