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hodyreva [135]
2 years ago
10

In the local frisbee league, teams have 7 members and each of the 4 teams takes turns hosting tournaments. At each tournament, e

ach team selects two members of that team to be on the tournament committee, except the host team, which selects three members. How many possible 9 member tournament committees are there?
Mathematics
1 answer:
mars1129 [50]2 years ago
8 0

There are 4 teams in total and each team has 7 members. One of the team will be the host team.

Tournament committee will be made from 3 members from the host team and 2 members from each of the three remaining teams. Selecting the members for tournament committee is a combinations problem. We have to select 3 members out 7 for host team and 2 members out of 7 from each of the remaining 3 teams.

So total number of possible 9 member tournament committees will be equal to:

7C3 \times 7C2 \times 7C2 \times 7C2\\ \\  = 324135

This is the case when a host team is fixed. Since any team can be the host team, there are 4 possible ways to select a host team. So the total number of possible 9 member tournament committee will be:

9 \times 324135\\ \\  =2917215

Therefore, there are 2917215 possible 9 member tournament committees

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Answer:

-2

Step-by-step explanation:

The leading coefficient in a polynomial is the coefficient that corresponds with the highest degree term. In this case, -2 is the leading coefficient since its term has x² as the highest degree of the quadratic.

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A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The
disa [49]

Answer:

The 90 % confidence limits are (-2.09, 8.09).

Since the calculated values do not lie in the critical region we accept our null hypothesis.

Step-by-step explanation:

The null and alternative hypothesis are given by

H0: σ₁²= σ₂² against Ha: σ₁² ≠ σ₂²

Confidence interval for the population mean difference is given by

(x`1- x`2) ± t √S²(1/n1 + 1/n2)

Where S ²= (n1-1)S₁² + S²₂(n2-1)/n1+n2-2

Critical value of t with n1+n2-2= 50+ 35-2= 83 will be -1.633

Now calculating

S ²=34* (12.8)²+ (14.6)²*49/83= 192.96

Now putting the values in the t- test

(75.1 -72.1) ± 1.633 √ 192.96(1/35 +1/50)

=3 ±  5.09

=-2.09, 8.09  is the 90 % confidence interval for the difference

The 90 % confidence limits are (-2.09, 8.09).

Since the calculated values do not lie in the critical region we accept our null hypothesis.

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3 years ago
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AlekseyPX

Answer: the answer IS G. MARK BRAILIEST PLS

Step-by-step explanation:

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2 years ago
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777dan777 [17]
(a) False. The number 7 is NOT an element of set B as 7 is not in the set B (2, 3, 4 and 0 are though)

(b) False. 3 is a member of set A. It is the third element listed in the set. So saying "3 is not a member of set A" is a false claim

(c) True. The value 0 is found in set B. It is the last element listed. 

Final Answer: Choice C) 
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3 years ago
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Ivahew [28]

Answer:

\sf Since \;\sqrt{\boxed{64}}=\boxed{8}\;and\;\sqrt{\boxed{81}}=\boxed{9}\; \textsf{it is known that $\sqrt{75}$ is between}\\\\\sf \boxed{8}\;and\;\boxed{9}\;.

Step-by-step explanation:

<u>Perfect squares</u>: 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ...

To find \sf \sqrt{75} , identify the perfect squares immediately <u>before</u> and <u>after</u> 75:

  • 64 and 81

\begin{aligned}\sf As\;\; 64 < 75 < 81\; & \implies \sf \sqrt{64} < \sqrt{75} < \sqrt{81}\\&\implies \sf \;\;\;\;\;8 < \sqrt{75} < 9 \end{aligned}

\sf Since \;\sqrt{\boxed{64}}=\boxed{8}\;and\;\sqrt{\boxed{81}}=\boxed{9}\; \textsf{it is known that $\sqrt{75}$ is between}\\\\\sf \boxed{8}\;and\;\boxed{9}\;.

See the attachment for the correct placement of \sf \sqrt{75} on the number line.

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9 months ago
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