five sharpshooters will each fire five shots at a target during a competition. The winner of the competition is the contestant who shot when the shortest average distance from the center of the target to the nearest 10th. The shots of the first four competitors land average distances of 4 cm, 4.5 cm, 3.5 cm, and 8 cm from the center of the target. The fifth sharpshooters first four shots are an average of 3.2 cm from the center of the target. What is the farthest distance from the center of the target, in centimeters, that the fifth sharpshooters final shot could be if she is the one who wins the competition?

Question

Annalise Hart

Mathematics

24 March, 13:07

five sharpshooters will each fire five shots at a target during a competition. The winner of the competition is the contestant who shot when the shortest average distance from the center of the target to the nearest 10th. The shots of the first four competitors land average distances of 4 cm, 4.5 cm, 3.5 cm, and 8 cm from the center of the target. The fifth sharpshooters first four shots are an average of 3.2 cm from the center of the target. What is the farthest distance from the center of the target, in centimeters, that the fifth sharpshooters final shot could be if she is the one who wins the competition?

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Answers (1)

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Areli Fox · Answer 1

13. 8 cm

Step-by-step explanation:

For the fifth sharpshooter to win, her average should be at most 3.4 since the best so far is 3.5.

so the equation would be

(3.2+x) / 5=3.4

3.2+x=3.4*5

3.2+x=17

x = 17 - 3.2

x = 13.8