In a video game, players form teams and work together to earn as many points as possible for their team. each team can have between 2 and 4 players. each player can score up to 20 points in each round of the game. han and three of his friends decided to form a team and play a round.

write an expression, an equation, or an inequality for each quantity described here. if you use a variable, specify what it represents.

the allowable number of players on a team

the number of points han's team earns in one round if every player earns a perfect score

the number of points han's team earns in one round if no players earn a perfect score

the number of players in a game with six teams of different sizes: two teams have 4
players each and the rest have 3 players each

the possible number of players in a game with eight teams

answer:

step-by-step explanation:

part a:

given that the game of clue involves 6 suspects, 6 weapons, and 9 rooms.

the number of ways that one of each is randomly chosen is given by:

^6c_1\times{ ^6c_1}\times{ ^9c_1}=6\times6\times9=324

therefore, the number of solutions possible is 324.

part b:

given that a players is randomly given three of the remaining cards, let s, w, and r be, respectively, the numbers of suspects, weapons, and rooms in the set of three cards given to a specified player.

the number of suspects, weapons, and rooms remaining respectively after the player observes his or her three cards are: 6 - s, 6 - w, and 9 - r.

let x denote the number of solutions that are possible after that player observes his or her three cards, then:

x={ ^{6-s}c_1}\times{ ^{6-w}c_1}\times{ ^{9-r}c_1}=(6-s)(6-w)(9-r)

therefore, x in terms of s, w, and r is given by x = (6 - s)(6 - w)(9 - r).

part c:

the expected value e(x) of a data set x_i with probabilities p(x_i) is given by e(x)=\sigma xp(x)

there are   ^{3+3-1}c_{3-1}={ ^5c_2}=10 possible combinations s, w and r. they are (3, 0, 0), (0, 3, 0), (0, 0, 3), (2, 1, 0), (0, 2, 1), (1, 0, 2), (2, 0, 1), (1, 2, 0), (0, 1, 2), (1, 1, 1)

thus the expected value is given by

e(x)=3\cdot6\cdot9p(3, 0, 0)+6\cdot3\cdot9p(0, 3, 0)+6\cdot6\cdot6p(0, 0, 3) \\ 4\cdot5\cdot9p(2, 1, 0)+6\cdot4\cdot8p(0, 2, 1)+5\cdot6\cdot7p(1, 0, 2)+4\cdot6\cdot8p(2, 0, 1) \\ +5\cdot4\cdot9p(1, 2, 0)+6\cdot5\cdot7p(0, 1, 2)+5\cdot5\cdot8(1, 1, 1) \\   \\ = \frac{1}{ ^{21}c_3} (162\cdot{ ^6c_3}\cdot{ ^6c_0}\cdot{ ^9c_0}+162\cdot{ ^6c_0}\cdot{ ^6c_3}\cdot{ ^9c_0}+216\cdot{ ^6c_0}\cdot{ ^6c_0}\cdot{ ^9c_3} \\ \\ +180\cdot{ ^6c_2}\cdot{ ^6c_1}\cdot{ ^9c_0}+192\cdot{ ^6c_0}\cdot{ ^6c_2}\cdot{ ^9c_1}

+210\cdot{ ^6c_1}\cdot{ ^6c_0}\cdot{ ^9c_2}+192\cdot{ ^6c_2}\cdot{ ^6c_0}\cdot{ ^9c_1}+180\cdot{ ^6c_1}\cdot{ ^6c_2}\cdot{ ^9c_0} \\   \\ +210\cdot{ ^6c_0}\cdot{ ^6c_1}\cdot{ ^9c_2}+200\cdot{ ^6c_1}\cdot{ ^6c_1}\cdot{ ^9c_1} \\   \\ =\frac{1}{1,330}(324\cdot20+216\cdot84+360\cdot90+384\cdot135+420\cdot216+200\cdot324) \\   \\ =\frac{1}{1,330}(6,480+18,144+32,400+51,840+90,720+64,800) \\   \\ =\frac{1}{1,330}(264,384) \\   \\ =\bold{198.78}

this doesn't make sense lol

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right