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The first-serve percentage of a tennis player in a match is normally distributed with a standard deviation of 4.3%. if a sample of 15 random matches of the player is taken, the mean first-serve percentage is found to be 26.4%. what is the margin of error of the sample mean? a. 0.086% b. 0.533% c. 1.11% d. 2.22%
Answers: 1
Solve for k:
2k(7-5k)+11=6k+3(k^2-1)
with work ....
2k(7-5k)+11=6k+3(k^2-1)
with work ....
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