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?

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

The problem statement is telling you Myra's speed at different points in her journey. First she is walking, then running, then stopped, then walking again. This suggests Myra's speed is the dependent variable. Often, problems involving speed or position use time as the independent variable. Here, that is suggested by the description, "... after a few minutes of walking, ...".

Myra's speed changes at the place where the spider web was, and again at the place where she met her friend (and when she reached school). Hence the independent variable could be position (distance), rather than time.