60 people attend a game night. every choo
cribbage, a four-player game. all 60 people are
end a game night. every chooses to play chess, a two-player game, or
our-player game. all 60 beople are playing either chess or cribbage.
part a
this table showing some possible combinations of the number of each type
of game being played.
chess games (x)
cribbage games (y)
-14
13.12=24-60 12
36
12
3 44 4 32-60=-20
+
91
15 14 40-60
part b
there are 3 more games of cribbage being played than games of chess being
played. how many of each game are being played? explain or show your reasoning.

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. release your mouse button when the item is place. if you change your mind, drag the item to the trashcan. click the trashcan to clear all your answers. using the technique in the model above, find the missing segments in this 30°-60°-90° right triangle. ab = 8 bc = 4 cd =

If a || b and b | y, then ? a) x || y. b) x | a. c) a | y. d) a || y.

Cone w has a radius of 8 cm and a height of 5 cm. square pyramid x has the same base area and height as cone w. paul and manuel disagree on how the volumes of cone w and square pyramid x are related. examine their arguments. which statement explains whose argument is correct and why? paul manuel the volume of square pyramid x is equal to the volume of cone w. this can be proven by finding the base area and volume of cone w, along with the volume of square pyramid x. the base area of cone w is π(r2) = π(82) = 200.96 cm2. the volume of cone w is one third(area of base)(h) = one third third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is one third(area of base)(h) = one third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is three times the volume of cone w. this can be proven by finding the base area and volume of cone w, along with the volume of square pyramid x. the base area of cone w is π(r2) = π(82) = 200.96 cm2. the volume of cone w is one third(area of base)(h) = one third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is (area of base)(h) = (200.96)(5) = 1,004.8 cm3. paul's argument is correct; manuel used the incorrect formula to find the volume of square pyramid x. paul's argument is correct; manuel used the incorrect base area to find the volume of square pyramid x. manuel's argument is correct; paul used the incorrect formula to find the volume of square pyramid x. manuel's argument is correct; paul used the incorrect base area to find the volume of square pyramid x.