Question 257913: In an especially brutal hockey game, 85 percent of the players lost a tooth, 80 percent broke a finger, 75 percent sprained an ankle, and 70 percent got a black eye. What is the smallest possible percent of players with all four injuries?
Found 2 solutions by Theo, CharlesG2:
Answer by Theo(13342)   (Show Source):
Answer by CharlesG2(834)  (Show Source):
You can put this solution on YOUR website! In an especially brutal hockey game, 85 percent of the players lost a tooth, 80 percent broke a finger, 75 percent sprained an ankle, and 70 percent got a black eye. What is the smallest possible percent of players with all four injuries?
Lets say there are 100 players, yes I know they do not play 100 players but it is easier number for coming up with ratios and percentages.
Actually: In National Hockey League (NHL) there are 20 players per team (18 regular plus 2 goal tenders) and with no penalties 6 per team on the ice at a time. So that is 40 players total with 12 total on the ice at a time.
so if 100 players
85 lost tooth
80 broke finger
75 sprained ankle
70 black eye
0.85*0.8*0.75*0.70=0.68*.75*.70=0.51*0.70=0.357 (multiply the percentages)
35.7 all 4 (yes know you can not have 0.7 people, that is just because of the percentage)
0.357*40 players both teams regulation NHL=14.28 people
think that would just be called 14 people, cause do not think anyone would agree to be cut into pieces but that would still be 14/40 or 35%
0.85*40=34 lost tooth
0.80*40=32 broke finger
0.75*40=30 sprained ankle
0.70*40=28 black eye
0.357*40 players both teams regulation NHL=14.28 people (or 14) with all 4
and if it was just 12 people:
0.85*12=10.2 or 10 lost tooth
0.80*12=9.6 or 9 broke finger
0.75*12=9 sprained ankle
0.70*12=8.4 or 8 black eye
0.357*12 players both teams regulation NHL=4.284 or 4 people with all 4
I do not think games like this happen very often, and the few games that do I imagine were quite intense to watch.
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