#1-2, can you explain as well on how to add or remove parentheses? 1.2.Consider the following expression.0*1 | 0(0*1)*What is the result when the order-of-precedence rules are used to add parentheses to the expression?O (((0*)1) | (0((0*)1)*))O ((0*)(1 | 0) ((0*)1)*)O ((0*)(1 | 0) (0*1))*O ((((0*)1) | 0)((0*1)*))O (((0)1) | (0(0*1))*)Consider the following expression.(xy)(((**)y)*) | (((yx) | y)(y*))What is the result when the order-of-precedence rules are used to eliminate the parentheses in the expression?O xyx*y* \ (yx \ yy*)xy(x*y)* | (yx | yy*)xy(x*y* | yx | y)y"xy(x*y)* | (yx | y)y*xyx*y* \ (yx \ y)y*

Note: As per guidelines I've provided answer for 1st problem only. Expression:…

Answered: Consider the following expression. 0*1…

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter8: Polynomials

Section8.1: Polynomials Over A Ring

Problem 13E

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2. The physicist Frank Benford noticed in the 1930s that, for many sets of numbers in the real world, the leading digit* of the number is more likely to be a small number than a larger one (i.e., 1’s and 2’s are more common than 8’s and 9’s). In fact, many wide-ranging number sets obey “Benford’s Law,” which says that 1’s are about 30.1% of the leading digits, 2’s are 17.6%, 3’s are 12.5%, 4’s are 9.7%, 5’s are 7.9%, 6’s are 6.7%, 7’s are 5.8%, 8’s are 5.1%, and 9’s are 4.6%. Benford’s Law has been used to identify accounting fraud and made-up scientific data. *leading digit is the first significant digit; e.g., “1” in the number 132, “2” in the number .0026.S uppose I suspect my research collaborator of faking a dataset that represents the price that people say is the average amount they pay each month in health care deductibles and copayments, rounded to the nearest dollar.10 50 15 11 412 80 48 13 4250 10 68 41 790 15 139 147 1433 61 145 7 406 99 50 27 7513 40 3 21 1475 203 148 27…
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Consider the following expression. 0*1 | 0(0*1)** What is the result when the order-of-precedence rules are used to add parentheses to the expression? O (((0*)1) | (0((0*)1)*)) ((0*)(1 | 0) ((0*)1)*) ((0*)(1 | 0) (0*1))* ((((0*)1) | 0) ((0*1)*)) O (((0*)1) | (0(0*1))*)

Consider the following expression. 01 | 0(01)** What is the result when the order-of-precedence rules are used to add parentheses to the expression? O (((0)1) | (0((0)1))) ((0)(1 | 0) ((0)1)) ((0)(1 | 0) (01))* ((((0)1) | 0) ((01))) O (((0)1) | (0(01)))