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Jill is calculating the standard deviation for her test grades. Mrs. Brown tells her that she did not take into account the curve on any of her grades and she needs to add 5 points to each grade. What is her new standard deviation? A) -2.204 Eliminate Reactivate B) 3.304 Eliminate Reactivate C) 5.804 Eliminate Reactivate D) 8.304
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