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sweet-ann [11.9K]
1 year ago
9

There are 15 players on a volleyball team. Only 6 players can be on the court for a game. How many different groups of players o

f 6 players can the coach make, if the position does not matter?
Mathematics
1 answer:
stealth61 [152]1 year ago
8 0

There are 5005 different groups

<h3>How to determine the number of groups?</h3>

The given parameters are:

Players, n = 15

Selected, r = 6

The number of groups is then calculated as:

Group = ^{n}C_r

This gives

Group = ^{15}C_6

Apply the combination formula

Group = \frac{15!}{9!6!}

Evaluate the expression

Group = 5005

Hence, there are 5005 different groups

Read more about combination at:

brainly.com/question/11732255

#SPJ1

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Answer:

10

Step-by-step explanation:

[1 -2]

[3 4]

We can obtain the determinant of the above matrix by doing the following:

Determinant =(1 × 4) – (3 × –2)

Determinant = 4 – – 6

Determinant = 4 + 6

Determinant = 10

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3 years ago
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Mariana [72]

Answer:

t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109      

The degrees of freedom are given by:

df=n-1=10-1=9  

And the p value would be given by:

p_v =P(t_{9}>1.109)=0.148  

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

Step-by-step explanation:

Information given

314 305 344 283 285 310​ 383​ 285​ 300​ 300

We can calculate the sample mean and deviation with the following formula:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s =\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}

\bar X=310.9 represent the sample mean      

s=31.09 represent the standard deviation for the sample      

n=10 sample size      

\mu_o =300 represent the value to test

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to verify if the true mean is greater than 300, the system of hypothesis would be:      

Null hypothesis:\mu \leq 300      

Alternative hypothesis:\mu > 300      

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)      

Replacing the info given we got:

t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109      

The degrees of freedom are given by:

df=n-1=10-1=9  

And the p value would be given by:

p_v =P(t_{9}>1.109)=0.148  

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

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The measure of central angle ABC is radians.
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Answer:

im not sure but maybe C

Step-by-step explanation:

sorry if this is wrong

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