Patrick B. answered 11/26/19
Math and computer tutor/teacher
100%/sqrt(2000) = 2.23606+%
Genesis P.
asked 11/26/19A polling organization conducts a poll by making a random survey of 2000 people. Estimate the margin of error at a confidence level of 95% using the rule of thumb for the margin of error. Round your answer as a percentage to one decimal place.
%
Patrick B. answered 11/26/19
Math and computer tutor/teacher
100%/sqrt(2000) = 2.23606+%
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