14th Edition

ISBN: 9781323907733

Author: Barnett

Publisher: INTER PEAR

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Textbook Question

Chapter 10.1, Problem 18E

Aims epidemic. One way to gauge the toll of the AIDS epidemic in Sub-Saharan Africa is to compare life expectancies with the figures that would have been projected in the absence of AIDS. Use the broken-line graphs shown to estimate the life expectancy of a child born in the year 2012 .What would the life expectancy of the same child be in the absence of AIDS? For which years of birth is the life expectancy less than 50 years? If there were no AIDS epidemic, for which years of birth would the life expectancy be less than 50 years.

Chapter 10.1, Problem 18E, Aims epidemic. One way to gauge the toll of the AIDS epidemic in Sub-Saharan Africa is to compare

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Chapter 10.1, Problem 17E

Chapter 10.1, Problem 19E

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7. [10 marks] Let G = (V,E) be a 3-connected graph. We prove that for every x, y, z Є V, there is a cycle in G on which x, y, and z all lie. (a) First prove that there are two internally disjoint xy-paths Po and P₁. (b) If z is on either Po or P₁, then combining Po and P₁ produces a cycle on which x, y, and z all lie. So assume that z is not on Po and not on P₁. Now prove that there are three paths Qo, Q1, and Q2 such that: ⚫each Qi starts at z; • each Qi ends at a vertex w; that is on Po or on P₁, where wo, w₁, and w₂ are distinct; the paths Qo, Q1, Q2 are disjoint from each other (except at the start vertex 2) and are disjoint from the paths Po and P₁ (except at the end vertices wo, W1, and w₂). (c) Use paths Po, P₁, Qo, Q1, and Q2 to prove that there is a cycle on which x, y, and z all lie. (To do this, notice that two of the w; must be on the same Pj.)

Chapter 10 Solutions

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Solution Summary: The author explains that the life expectancy of a child born in the year 2012 is nearly 53.

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