2021 Mathematics Final Exam With Answers (40 Solved Questions)

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111/34Module 8: Final ExamDueNo due datePoints200Questions40Time Limit400 MinutesAttempt HistoryAttemptTimeScoreLATESTAttempt 1111 minutes155 out of 200Correct answers are hidden.Score for this quiz:155out of 200Submitted Nov 1 at9:56pmThis attempt took 111 minutes.5/ 5ptsQuestion 1

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112/34Consider the following graph:(Section 4.4)Alt text: A graph with 4 vertices and 7 edges connecting them.What is the minimum number of edges that must be added in order for thegraph to contain an Euler path?1234

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113/34Correct. Note that: Adding just one edge that connects two verticeseach with odd degree, will leave the graph with only two vertices ofodd degree. Hence, the new graph will contain an Euler path.5/ 5ptsQuestion 2A high school math department offers a total of 5 classes: MTH01,MTH02, MTH03, MTH04, and MTH05. However, the following coursescannot be taught at the same time due to staffing limits:• MTH01 conflicts with MTH02, MTH05• MTH02 conflicts with MTH01, MTH03• MTH03 conflicts with MTH02, MTH04• MTH04 conflicts with MTH03, MTH05• MTH05 conflicts with MTH01, MTH04What graph could we use to model this situation?A graph with 5 vertices that represent the classes which have a conflictand edges connecting all classes.A graph with 5 vertices that represent the classes and edges connectingclasses which have a conflict.A graph with 5 vertices that represent the classes and edges connecting allclasses.A graph with 5 vertices that represent the classes and edges connectingtwo of the classes.

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114/34Correct. Note that: We could create a graph with 5 vertices thatrepresent the classes, and edges connecting classes which have aconflict.5/ 5ptsQuestion 3Which of the following would be the correct way to begin a proof bycontraposition of the statement, “If there is snow, then it is winter”?Suppose it is winter.Suppose there is snow.Suppose it is not winter.Suppose there is no snow.Correct. Note that: In a proof by contraposition, we begin byassuming the negation of the conclusion. In this case, this wouldbe "it is not winter."5/ 5ptsQuestion 4Consider the statement, “For all integersaandb, if the product ofaandbis even, thenais even orbis even.” What would be the beginning of aproof by contradiction of this statement?Assume that the product ofaandbis odd, andaandbare both odd.

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115/34Assume that the product ofaandbis even, andaandbare both even.Assume that the product ofaandbis even, andaandbare both odd.Assume that the product ofaandbis odd, andais even orbis odd.Correct. Note that: We must assume the hypothesis and thenegation of the conclusion. Hence, we begin as follows: Assumethat the product ofaandbis even, andaandbare both odd.5/ 5ptsQuestion 5Consider the statement, “For all integersaandb, if the product ofaandbis even, thenais even orbis even.” What would be the beginning of aproof by contrapositive of this statement?Assume that eitheraorbare odd. We must show thatabis odd.Assume thataandbare both odd. We must show thatabis even.Assume thataandbare both even. We must show thatabis even.Assume thataandbare both odd. We must show thatabis odd.Correct. Note that: A proof by contraposition would begin asfollows: Assume thataandbare both odd. We must show thatabis odd.0/ 5ptsQuestion 6IncorrectIncorrect

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116/34Which of the following is logically equivalent to the statement, “If a book isreturned late, then there is a fee?”If there is no fee, then the book is returned late.If there is no fee, then the book is not returned late.If there is a fee, then the book is not returned late.If there is a fee, then the book is returned late.Actually, "If there is no fee, then the book is not returned late" isthe contrapositive of the statement; so, we know it is logicallyequivalent.0/ 5ptsQuestion 7IncorrectIncorrectA high school math department offers a total of 5 classes: MTH01,MTH02, MTH03, MTH04, and MTH05. However, the following coursescannot be taught at the same time due to staffing limits:• MTH01 conflicts with MTH02, MTH05• MTH02 conflicts with MTH01, MTH03• MTH03 conflicts with MTH02, MTH04• MTH04 conflicts with MTH03, MTH05• MTH05 conflicts with MTH01, MTH04We want to know the minimum number of time slots needed to teachthese classes. What information about the graph would answer thisquestion?number of verticesnumber of edges

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117/34chromatic indexchromatic numberActually, the chromatic number of this graph would tell us howmany time slots are needed to teach these classes.5/ 5ptsQuestion 8Consider the statement, “For all integersaandb, if the product ofaandbis even, thenais even orbis even.” What would be the beginning of adirect proof of this statement?Assume thatais even orbis even. Thena or b= 2kfor some integerk.We must show that eitherais even orbis even, so it suffices to show oneor the other.Assume thatabis even. Thenab= 2k+1 for some integerk.We mustshow that eitherais even orbis even, so it suffices to show one or theother.Assume thatabis even. Thenab= 2kfor some integerk.We must showthat eitherais even orbis even, so it suffices to show one or the other.Assume thatabis even. Thenab= 2kfor some integerk.We must showthatais even andbis even.

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118/34Correct. Note that: A direct proof would begin as follows: Assumethatabis even. Thenab= 2kfor some integerk.We must showthat eitherais even orbis even, so it suffices to show one or theother.0/ 5ptsQuestion 9IncorrectIncorrectA high school math department offers a total of 5 classes: MTH01,MTH02, MTH03, MTH04, and MTH05. However, the following coursescannot be taught at the same time due to staffing limits:• MTH01 conflicts with MTH02, MTH05• MTH02 conflicts with MTH01, MTH03• MTH03 conflicts with MTH02, MTH04• MTH04 conflicts with MTH03, MTH05• MTH05 conflicts with MTH01, MTH04What is the minimum number of time slots needed to teach theseclasses?3456Actually, by trial and error, we see that the circuit can be coloredwith 3 colors. Hence, a minimum of 3 time slots are required.5/ 5ptsQuestion 10

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119/34Which of the following graphs has exactly 5 edges? (Check all that apply.)(Section 4.1)Alt text: A graph with five vertices that are all connected to each other byedges.(Section 4 1)

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1110/34(Section 4.1)Alt text: A graph with 6 vertices and 5 edges connecting them to form apath.*c.(Section 4.1)Alt text: A graph with five connected vertices that form a pentagon.d.

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