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:

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:

Therefore, there are 2917215 possible 9 member tournament committees
Answer:
All are rational.
Step-by-step explanation:
1)
=1/3/3/5
=1/3×5/3
=5/9
5/9 is non-terminating and recurring number so it is rational.
2)
4/3√9/5
=4/3*3/5
=4/5
4/5 is a terminating number it is rational.
3)
2*√16/√4
=2*4/2
=4
4 is a natural/whole number it is also rational.
Note<u>:</u><u>i</u><u>f</u><u> </u><u>y</u><u>o</u><u>u</u><u> </u><u>n</u><u>e</u><u>e</u><u>d</u><u> </u><u>t</u><u>o</u><u> </u><u>a</u><u>s</u><u>k</u><u> </u><u>a</u><u>n</u><u>y</u><u> </u><u>questions</u><u> </u><u>please</u><u> </u><u>l</u><u>e</u><u>t</u><u> </u><u>m</u><u>e</u><u> </u><u>k</u><u>n</u><u>o</u><u>w</u><u>.</u>
Step-by-step explanation:
Geometric series.
Month 3.
100(1+\frac{0.02}{12})^2 + 100(1+\frac{0.02}{12})+100
Month 4.
100( 1 + \frac{0.02}{12})^3 + 100( 1 + \frac{0.02}{12})^2+100(1+\frac{0.02}{12})+100
Month 5.
100(1 + \frac{0.02}{12})^4 + 100(1 + \frac{0.02}{12})^3 + 100(1 + \frac{0.02}{12})^2 +100( 1 + \frac{0.02}{12}) + 100
Answer:
i think if it were me i whould divide 300 dide 15
Step-by-step explanation: