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denis23 [38]
3 years ago
14

How many different ways can a coach select the captain and co-captain of a team from a group of 20 people?

Mathematics
2 answers:
sergij07 [2.7K]3 years ago
5 0
The coach can select the captain in 20 ways (i.e. from all the available 20 players), and he can select the co-captain in 19 ways (i.e. after selecting the captain, there will be 19 people left).
Therefore, the coach can select the captain and the co-captain in 20 x 19 = 380 ways.
Thepotemich [5.8K]3 years ago
3 0

Answer:

Step-by-step explanation:

1. Number of peoples available to choose from = 20

\text{Ways of selecting the captain from 20 people = }_{1}^{20}\textrm {C}=20

After selecting the captain number of people left among which co captain can be selected = 20 - 1 = 19

\text{Ways of selecting the co captain from 19 people = }_{1}^{19}\textrm {C}=19

Hence, Number of ways a coach select the captain and co-captain of a team from a group of 20 people = 20 × 19 = 380

Therefore, The correct option is C. 380

2. Total number of players = 7

Number of players to be selected = 5

\text{So, number of ways of selecting 5 players out of 7 = }_{5}^{7}\textrm {C}=21

Therefore, The correct option is A. 21

3. Total number of outcomes = 20

The digits 1, 2, 3, 4, 5, 6 represent the student who make a C

Number of students who make a C at least 3 times = 13

\bf\textbf{Required Probability = }\frac{13}{20}

Therefore, The correct option is D

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Given the probability distributions shown to the​ right, complete the following parts.
Elan Coil [88]

Answer:

a) Expected Value for distribution A, E(X) = 3.020

Expected Value for distribution B, E(X) = 0.980

b) Standard deviation of distribution A = 1.157

Standard deviation of distribution B = 1.157

c) In distribution A, the bigger values of x have a higher probability of occurring than the values of distribution B (whose smaller values of x have a higher chance of occurring, hence, the expected value for distribution A is more than that of distribution B.

But according to the corresponding distribution of values, the two distributions have the same exact spread, A in ascending order (with higher values with bigger probability) and B in descending order (lower values have higher probabilities). But the same spread regardless, hence, the standard deviation which shows how data values spread around the mean (centre point) of a distribution is the same for the two distributions.

Step-by-step explanation:

Expected values is given as

E(X) = Σ xᵢpᵢ

where xᵢ = each possible sample space

pᵢ = P(X=xᵢ) = probability of each sample space occurring.

Distributions A and B is given by

X P(X) X P(x)

0 0.04 0 0.47

1 0.09 1 0.25

2 0.15 2 0.15

3 0.25 3 0.09

4 0.47 4 0.04

For distribution A

E(X) = Σ xᵢpᵢ = (0×0.04) + (1×0.09) + (2×0.15) + (3×0.25) + (4×0.47) = 3.02

For distribution B

E(X) = Σ xᵢpᵢ = (0×0.47) + (1×0.25) + (2×0.15) + (3×0.09) + (4×0.04) = 0.98

b) Standard deviation = √(variance)

But Variance is given by

Variance = Var(X) = Σx²p − μ²

where μ = E(X)

For distribution A

Σx²p = (0²×0.04) + (1²×0.09) + (2²×0.15) + (3²×0.25) + (4²×0.47) = 10.46

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For distribution B

Σx²p = (0²×0.47) + (1²×0.25) + (2²×0.15) + (3²×0.09) + (4²×0.04) = 2.30

Variance = Var(X) = 2.30 - 0.98² = 1.3396

Standard deviation = √(1.3396) = 1.157

c) In distribution A, the bigger values of x have a higher probability of occurring than the values of distribution B (whose smaller values of x have a higher chance of occurring, hence, the expected value for distribution A is more than that of distribution B.

But according to the corresponding distribution of values, the two distributions have the same exact spread, A in ascending order (with higher values with bigger probability) and B in descending order (lower values have higher probabilities). But the same spread regardless, hence, the standard deviation which shows how data values spread around the mean (centre point) of a distribution is the same for the two distributions.

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