Mathematical Applications For The Management, Life, And Social Sciences, 11th Edition Solution Manual

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Chapter 0: Algebraic Concepts1Exercises 0.1 __________________________________________________________________1.12{1, 2, 3, 4,...}2.5{x:xis a natural number greater than 5}3.6{1, 2, 3, 4, 5}4.35.{1, 2, 3, 4, 5, 6, 7}6.{7, 8, 9}7.{x: xis a natural number greater than 2 and lessthan 8}8.{x:xis a natural number greater than 6}9.A sinceis a subset of every set.ABsince every element ofAis an element ofB.BBsince a set is always a subset of itself.10.A sinceis a subset of every set.ABsince every element ofAis an element ofB.BBsince a set is always a subset of itself.11.No.cAbut.cB12.No.12Abut12.B13.DCsince every element ofDis an elementofC.14.EFsince every element ofEis an elementofF.15.ABandBA. (AlsoAB.)16.and. (Also.)DFFDDF17.Yes.ABandBA. Thus,AB.18.AD19.No.DEbecause4Eand4.D20.F=G21.AandBare disjoint since they have no elementsin common.BandDare disjoint since they haveno elements in common.CandDare disjoint.22.23.{4, 6}ABsince 4 and 6 are elements of eachset.24.{ ,,}ABa d esincea,d, andeare elementsof each set.25.AB=since they have no commonelements.26.{3}AB27.{1, 2, 3, 4, 5}AB28.{ ,,,,, ,,}ABa b c d e i o u29.{1, 2, 3, 4}ABor.ABB30.AB{x:xis a natural number not equal to5}For problems 31 - 42, we haveU= {1, 2, 3, . . . , 9, 10}.31.{4, 6, 9, 10}A since these are the onlyelements inUthat are not elements ofA.32.{1, 2, 5, 6, 7, 9}B since these are the only elements inUthat arenot elements ofB.33.{1, 2, 5, 6, 7, 9}B {1, 2, 5,7}AB34.{4, 6, 9, 10}{1, 2, 5, 6, 7, 9}{6, 9}ABAB  35.1, 2, 3, 4, 5, 7, 8, 10AB(){6, 9}AB36.{3, 8}(){1, 2, 4, 5, 6, 7, 9, 10}ABAB37.{4, 6, 9, 10}A {1, 2, 5, 6, 7, 9}B {1, 2, 4, 5, 6, 7, 9, 10}AB

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Chapter 0: Algebraic Concepts238.{4, 6, 9, 10}{3, 4, 8, 10}{3, 4, 6, 8, 9, 10}(){1, 2, 5, 7}ABABAB  39.{1, 2, 5, 6, 7, 9}B {1, 3, 5, 7, 9}C 1, 2, 3, 5, 7, 81, 2, 5, 6, 7, 91, 2, 5, 7AB1, 2, 3, 5, 7, 9ABC40.{1, 3, 5, 8, 7, 2}{1, 2, 5, 6, 7, 9}{1, 3, 5, 7, 9}{1, 2, 3, 5, 6, 7, 9}(){1, 2, 3, 5, 7}ABCBCABC  41.{1, 2, 5, 6, 7, 9}B 1, 2, 3, 5, 7, 81, 2, 5, 6, 7, 91, 2, 5, 7AB3, 4, 6, 8, 9, 102, 4, 6, 8, 104, 6, 8, 10ABC42.{2, 3, 4, 6, 8, 10}(){2, 3, 8}BCABCFor problems 43 - 46, we haveU= {1, 2, 3, . . . , 8, 9}.43.A–B= {1, 3, 7, 9}–{3, 5, 8, 9} = {1, 7}44.A–B= {1, 2, 3, 6, 9}–{1, 4, 5, 6, 7} = {2, 3, 9}45.A–B= {2, 1, 5}–{1, 2, 3, 4, 5, 6} =or 46.A–B= {1, 2, 3, 4, 5}–{7, 8, 9} = {1, 2, 3, 4, 5}47.a.L= {2000, 2001, 2004, 2005, 2006, 2007, 2010, 2011, 2012}H= {2000, 2001, 2006, 2007, 2008, 2010, 2011, 2012}C= {2001, 2002, 2003, 2008, 2009}b.noc.Cis the set of all years when the percentage change from low to high was 35% or less.d.H= {2002, 2003, 2004, 2005, 2009}C= {2000, 2004, 2005, 2006, 2007, 2010, 2011, 2012}HC= {2000, 2002, 2003, 2004, 2005, 2006, 2007, 2009, 2010, 2011, 2012}.HCis the set ofyears when the high was less than or equal to 11,000 or the percent change was less than or equal to 35%.e.L= {2002, 2003, 2008, 2009}L C= {2002, 2003, 2008, 2009}.L Cis the set of years when the low was less than or equal to 8,000 and the percent change was morethan 35%.48.a.A= {O, L, P}B= {L, P}C= {O, M, P}b.BAc.{O, P};ACthis is the set of cities with at least 2,000,000 jobs in 2000 or 2025 and projected annualgrowth rates of at least 2.5%.d.Bis the set of cities with fewer than 1,500,000 jobs in 2000.49.a.From the table, there are 100 white Republicans and 30 non-white Republicans who favor nationalhealth care, for a total of 130.b.From the table, there are 350 + 40 Republicans, and 250 + 200 Democrats who favor national health care,for a total of 840.c.From the table, there are 350 white Republicans, and 150 white Democrats and 20 non-whites who opposenational health care, for a total of 520.

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Chapter 0: Algebraic Concepts350.a.From the table, 250 white Republicans and 150 white Democrats oppose national health care, for a total of400.b.From the table, there are 750 whites and there are 20 non-whites who oppose national health care. The totalof this group is 770.c.From the table, there are 200 non-white Democrats who favor national health care.51.a.The key to solving this problem is to work from "the inside out". Thereare 40 aides inEF. This leaves 65–40 = 25 aides who speak Englishbut do not speak French. Also we have 60–40 = 20 aides who speakFrench but do not speak English. Thus there are 40 + 25 + 20 = 85 aideswho speak English or French. This means there are 15 aides who do notspeak English or French.b.From the Venn diagramEFhas 40 aides.c.From the Venn diagramEFhas 85 aides.d.From the Venn diagramEFhas 25 aides.52.a.There are 14 advertisers in the intersection of the sets. Since 30 advertised inThese TimesandU.S. Newsand we already have 14 in the center, 16 advertised inThese TimesandU.S. Newsand not inWorld. Since26 advertised inWorldandU.S. Newsand we already have 14 in the center, 12 advertised inWorldandU.S. Newsand not inThese Times. Since 27 advertised inWorldandThese Timesand we already have 14in the middle, 13 advertised inWorldandThese Timesand not inU.S. News. 60 advertised inThese Timesand we have already accounted for 43, so 17 advertised inThese Timesonly. 52 advertised inU.S. Newsand we have already accounted for 42, so 10 advertised inU.S. Newsonly. 50 advertised inWorldand wehave already accounted for 39, so 11 advertised inWorldonly.b.In the union of the 3 publications we have 10 + 16 + 17 + 14 + 12 + 13+ 11 = 93 advertisers. Thus, there are 100–93 = 7 who advertised innone of these publications.c.There are 17 advertisers in theThese Timescircle that are not in anintersection.d.In the union ofU.S. NewsandThese Timeswe have 10 + 12 + 16 + 14+ 17 + 13 = 82 advertisers.53.Since 12 students takeMandEbut notFA, and 15 takeMandE, 3 take all three classes. Since 9 students takeMandFAand we have already counted 3, there are 6 takingMandFAwhich are not takingE. Since 4 studentstakeEandFAand we have already counted 3, there is only 1 takingEandFAbut not takingMalso. Since 20students takeEand we already have 16 enrolled inE, this leaves 4 taking onlyE. Since 42 students takeFAandwe already have 10 enrolled inFA, this leaves 32 taking onlyFA. Since 38students takeMand we already have 21 enrolled inM, this leaves 17 takingonlyM.a.In the union of the 3 courses we have 17 + 12 + 3 + 6 + 32 + 1 + 4 = 75students enrolled. Thus, there are 100–75 = 25 students who are notenrolled in any of these courses.b.InMEwe have 17 + 12 + 3 + 6 + 1 + 4 = 43 enrolled.c.We have 17 + 32 + 4 = 53 students enrolled in exactly one of the courses.UEF25402015UU.S. News10121311These TimesWorld1617147UMath171214Fine ArtsEconomics632325

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Chapter 0: Algebraic Concepts454.Start by filling in the parts of the diagram for AL, since we have more information about it. 21 liked AL only.Since 30 liked AL but not PT, 9 liked AL or PT exclusively. 25 liked PT or AL but not DH, and 63 liked AL.That leaves 63–(21 + 25 + 9) = 8 in the intersection of all 3. Since 18 likedPT and DH, only 10 liked PT and DH but not AL. Since 27 liked DH, 27–(9+ 8 + 10) = 0 liked DH only. And since 58 liked PT, 58–(25 + 8 + 10) = 15liked PT only.a.The number of students that liked PT or DH is25 + 15 + 9 + 8 + 10 + 0 = 67.b.The number that liked all three is 8.c.The number that liked only DH is 0.55.a.andb.c.: 34%;A: 9%B;: 38%O;: 3%AB;: 7%O;: 6%A;: 2%B;: 1%ABExercises 0.2__________________________________________________________________1.a.Note that11010 , whereisirrational and110is rational. The productof a rational number other than 0 and anirrational number is an irrational number.b.–9 is rational and an integer.c.93331. This is a natural number, aninteger, and a rational number.d.Division by zero is meaningless.2.a.006is rational and an integer.b.rationalc.rationald.rational3.a.Commutativeb.Distributivec.Associatived.Additive Identity4.a.Multiplicative Identityb.Additive Inversec.Multiplicative Inversed.Commutative5.–6 < 06.2 >–207.–14 <–38.3.14 9.110.3330.33333310.11532611.|3 || 5 ||35 |  12.|93 ||9 || 3 | (12 = 12)13.22310 232092011   14.2( 3)10 292029UAL219100PTDH2515822UAL219100PTDH2515822UAABOBABABRhABO

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Chapter 0: Algebraic Concepts515.242448422216.22(42)6361822217.16( 4)1642028( 2)8210  18.( 5)( 3)( 2)(3)15( 6)15692772137   19.| 52 ||7 || 3 ||7 |374| 52 || 3 |33   20.22| 3| 411||| 3|7 ||| 259 || 53 || 37 ||16 ||4 |1641164     21.22( 3)2 3696693443342322.2264( 3)( 2)36( 12)( 2)63646643624691234   23.22452 31656171751611115424.2232(52)32 3314433( 2)23 25.The entire line26.The interval notation corresponding to0xis[0,).27.(1, 3]; half-open interval28.[–4, 3]; closed interval29.(2, 10); open interval30.[2,);half-open interval31.35x32.x>–233.x> 434.05x35.(, 4)( 3,)( 3, 4)   36.[–4, 17)[–20, 10] = [–4, 10]37.x> 4 and0x=(4,)38.x< 10 andx<–1 isx<–1 or(,1) .39.[0,)[ 1, 5][ 1,)   40.(, 4)(0, 2) =(, 4)41.(,0)(7,)42.x> 4 andx< 0The intersection is the empty set.43.–0.00003858544.0.40478702545.9122.38747112345310246810126204123456102123456102123456102123456102123456107123456102

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Chapter 0: Algebraic Concepts646.11.8059162147.6250025003240.1845090.771561[(1.1 )1]Equation (2) gives20.00454 130.126 130.2712.68 billiony48.1591.71265249.a.$300.00$788.91$1088.91b.Federal withholding= 0.25(1088.91–54.45) = $258.62c.Retirement: 0.05(1088.91) = $54.45State tax = Retirement = $54.45Local tax = 0.01(1088.91) = $10.89Federal tax = $258.62 (fromb.above)Social Security and Medicare taxd= 0.0765(1088.91) = $83.30Total Withholding = $461.71Take-home = 1088.91–461.71 = $627.2050.a.2010200010tb.25.03 10100 101380E$2883 billionc.2201520001550.3 15100 151380$4011.75 billiontE51.a.Equation (1) is more accurate.dEquation (1) givesd0.207 130.0003702.69 billionyb.For 2018, Equation (1) gives0.207 180.0003703.73 billionyFor 2018, Equation (2) gives20.00454 180.126 180.2714.01 billiony52.a.2.31 10.531.2655.515 inchesUpper: 1.05 55.51558.29 inchesLower: 0.95 55.51552.74 inches52.7458.29HHb.2.31 5.7531.2644.5425 inchesUpper: 1.05 44.542546.77 inchesLower: 0.95 44.542542.32 inches42.3246.77HH53.a.$82, 401171,850;$171,851$373, 650;$373, 650IIIb.$4681.25 for$34, 000$16, 781.25 for$82, 400TITIc.4681.25, 16,781.25Exercises 0.3 __________________________________________________________________1.4444442562.351 5 5 5125   3.621 2 2 2 2 2 264   4.522222232 5.22113936.11667.2333912224 8.33322832739.1.242.0736xy10.3.7350.653xy11.1.550.1316872428xy12.0.897.450580597xy

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Chapter 0: Algebraic Concepts713.535 38666614.4242 1788888 15.88 9191011010101016.88353777717.474( 7)33 ( 3)03339999991999   18.4444 1323123555555555 19.333 3933320.23(3) (2)6222 21.2223932422.44425552223.33331111xxxx     24.441xx25.202211xxyzxyy26.102221114144x yyy 27.343 47xxxx28.55 16aaaa29.5353221xxxxx 30.525( 2)771yyyyy  31.88444xxxx32.55( 1)5 161aaaaa 33.557127yyyy 34.33 ( 4)3414yyyyyy   35.343 412xxx36.323 (2)661()yyyy 37.222()xyx y38.3333(2)28mmm39.44545 4205221616xxxx40.333333 3988512512()aaaa41.84248442216xxyx yy42.335355( 3)3151532( 32)1( 32)132768132768xxxxx  

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Chapter 0: Algebraic Concepts843.32543524222282161616aba baba bab    44.21312(3)1 (1)122232661166m ymymymymymy        45.221222222222yxxxyxxxyxy46.323254543254828822824ab cab ca ba babcabca b47.339269696111xxyyxyx y48.3223( 2)( 3)63633363()1xxxxyxyyyyx y49.33321426418124306426abcba caca bca cbb50.221401 440( 10)24102442(1/ 2)xyxyx y  5302252302(5)(2)(30)(2)210601060(16)(16)()()1(16)1256256xyxyxyxyxy51.a.222222421111122(2 )(2 )42xxxxxxxb.22222242111114416222xxxxxxxc.222242111222xxxxxd.222222211222482xxxxxx52.a.12121 22222222222xxxxx  344128xxb.121 1222424212224xxxxx  c.122111222222222222xxxxxx1222422xx   d.211( 2)( 1)12222(2)22222xxxxxx1 122124x 53.11xx54.221xx55.3333228xxx56.2222(3 )39xxx

Mathematical Applications For The Management, Life, And Social Sciences, 11th Edition Solution Manual - Page 10 preview image

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Chapter 0: Algebraic Concepts957.222111144(4)xxx58.444331322(2)xxx59.33331282xxx 60.22222()13993xxxx61.551200,0.12,5(1)1200(10.12)1200(1.12)$2114.81nPinSPi–2114.81–1200$914.81ISP62.1800P,0.10i,7n77(1)1800(10.10)1800(1.10)1800(1.9487171)$3507.69nSPi3507.691800$1707.69ISP63.665000,0.115,6(1)5000(10.115)5000(1.115)$9607.70nPinSPi–9607.70 – 5000$4607.70ISP64.800P,0.105,i20n2020(1)800(10.105)800(1.105)5892.99nSPi5892.99800$5092.99ISP65.S= 15,000,n= 6,i= 0.11566115, 000 10.11515, 000 1.115$7806.24nPSi66.80,000,20,0.105Sni2020(1)80, 000(10.105)80, 000(1.105)$10,860.37nPSi67.492.4 1.070tI58Year198020002008.-value204048.Income$1905$7373$12, 669(in billions).492.4 1.070$24,922 billiontIabc68.a.201920109tb.90.012 1.751.8 billion cubic feetyc.120.012 1.759.9 billion cubic feety

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Chapter 0: Algebraic Concepts1069.1095110.12 1.212ty.Year199020032012value102332Predicted number of4429761072endangered speciestab.Year 2020:40t;4010951090110.12 1.212yIncrease between 2007 and 2020 is1090103753speciesc.Two possibilities might be more environmental protections and the fact that there are only a limitednumber of species.d.There are only a limited number of species. Also, below some threshold level the ecological balance mightbe lost, perhaps resulting in an environmental catastrophe (which the equation could not predict).To findthe upper limit, which is 1095, computeyfor larget-values:Year204021002200value60120220Predicted number of1094.910951095endangered speciest70.249.611.915 1.028tp.Year198020002020value305070U.S. population(age 20 - 64)135.92168.49195.44in millionstab.Year 2025:75t;75249.6201.07 million11.915 1.028pYear 2045:95t;95249.6219.15 million11.915 1.028pThe increase from 2025 and 2045 is predicted to be219.15201.0718.08million. This is less than thepredicted 28.4 million increase from 2000 to 2020.c.It is reasonable for a formula such as this to have an upper limit that cannot be exceeded because there arelimited resources and space. To find the upper limit, which is 249.6 million, computepfor larget-values:Year215023502450value200400500U.S. population(age 20 - 64)247.7249.6249.6in millionst71.738.1 1.065tHa.10tcorresponds to 2000.

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Chapter 0: Algebraic Concepts11b.10738.1 1.065$1385.5 billionHc.20738.1 1.065$2600.8 billionHd.28738.1 1.065$4304.3 billionHExercises 0.4 __________________________________________________________________1.a.Since21625639we have256163935. 3b.1.441.22.a.5553333/5335321 3232(32)32(2)8      b.454161048576The square root of anegative number is not real.3.a.33/434161628b.33/2( 16)16The square root of anegative number is not real.4.a.1/31/33112727327   b.33 5353232285.22/32/32382727392788246.33/234428993277.a.22/3236464416b.2/32/323211646464111648.a.2/32/32231111641664464b.222/336464416   9.4/949(6.12)6.122.23710.1/12124.964.961.142811.33/2mm12.355/3xx13.1/4425252/45/41/25/4m nm nmnmn14.533/5xx15.1222xx16.1441212xx17.67/67xx18.11/5115yy19.5/45/4451111444xxx 20.35/35351xxx 21.1/41/2(1/4)(1/2)3/4yyyy22.2/31/5(2/3)(1/5)(10/15)(3/15)13/15xxxxx23.3/44(3/4)(16/4)19/4zzzz

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Chapter 0: Algebraic Concepts1224.2/32( 2/3)2( 2/3)(6/3)4/3xxxxx25.3/21( 3/2)(2/2)5/25/21yyyyy26.25/32(5/3)( 6/3)(5/3)1/31/31zzzzzz 27.1123333323xxxxx28.1/2( 1/2)( 3/2)( 1/2)(3/2)2/23/2xxxxxx 29.5/2( 5/2)( 2/5)( 25/10)(4/10)2/5yyyy 21/1021/101yy30.4/9(4/9)(1/12)(16/36)(3/36)13/361/12xxxxx31.2/3 3/4(2/3)(3/4)2/41/2()xxxx32.4/5 3(4/5)(3)12/5()xxx33.1/2211()xxx34.2/32/5( 2/3)( 2/5)4/15()xxx35.42648xx36.36363233364644x yxyx y 37.4544442212864264282x yx yyxyyx yy38.33358583333222222354543232x zxzxxzzxzx z39.8563223336322333222340858525x yx yx yxyx yx yx y40.552323242x yxyxxy242xxy41.325252212336366x yx yx yxyx yx42.32242423333322333316348482626x yx yx yxyxxyxxy43.53242223263289747372742x yx yx yxyxyx yxyxyx yx44.10171811181195951030300300103103xzxzxzxzxzzx zz45.3122105212429327x yx yxyxy46.747454172161725416228228250250125189181259553553xy zxy zy zxyxxyyzxyyzxy zyx47.49544482417321621813162a bbbbbaaa

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Chapter 0: Algebraic Concepts1348.3343234333232322331616128812822x yxyx yyyxyxy49.9991()9119xxxAAAAxx50.20201()201120xxBBBBxx51./77/71177xxxRRRRxx52.33 1/213/213 /21(())()31223xxxxTTTTTTTTxx53.232363333354.5858402 10108848886455.22m xmmxmxxxxxmx56.333222255552244x wxx wxx wxxxxwxxwx w57.3332223333333543233m xmmxmxmxxxxxxx322mxx58.23323323434442252348444y zmx y zmx y zmxyzy zy zy z59.2/32/3322212333xxx 60.3/43/43/44322223333xxxx 61.1/23/2333xxx xx62.1/21/3(1/2)(1/3)(3/6)(2/6)3xxxxxx5/6x63.1/23322xx64.31/33444333xxx65.1/21/21111222xxx66.3/23/23111222xxx67.a.17/217178.510102RIb.9.0101,000,000,000Ic.9.02.120116.919891010125.910II68.a.10/101/1010(10)10DDDb.10321101584.89I

Mathematical Applications For The Management, Life, And Social Sciences, 11th Edition Solution Manual - Page 15 preview image

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Chapter 0: Algebraic Concepts14c.10140101014032108210321(108)(1/10)10.8101010101010106.31 10II69.a.510001100rSb.56.610001$1173.26100S70.a.10021210.21, so29100Lxb.0.2129 11578.5 yearsL71.a.10013/100130.9240.924Pttb.YearPopulation200551.13902010101.24642045451.51562050501.5365tChange from2005 to 2010 :0.1074 billionChange from2045 to 2050 : 0.0209 billionBy 2045 and 2050 the population is much larger than earlier in the 21stcentury, and there is a limitednumber of people that any land can support—in terms of both space and food.72.a.500.3838/10019/501914.3214.3214.3214.32pttttb.YearPercent of Roads Paved19702044.719803052.120005063.320106067.9tChange from1970 to 1980 : 7.4%Change from2000 to 2010 : 4.6%The equation estimates a greater percent change from 1970 to 1980 than from 2000 to 2010. There werefewer roads left to be paved from 2000 to 2010.c.When at-value makes100%,pyou can be certain that the equation is no longer valid since you cannotpave more than 100% of the roads.73.25,10,kt098q/010/252/5(2)98(2)98(2)74 kgt kqq74.5600,10,000,kt040q/010,000/560025/14(2)40(2)40(2)11.6 gt kqq75.0.03 1002.530, 000 2.5htPP0.330, 000 2.539, 491

Mathematical Applications For The Management, Life, And Social Sciences, 11th Edition Solution Manual - Page 16 preview image

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Chapter 0: Algebraic Concepts1576.10x0.10.1 1012000 22000 22000 212000 2$1000xS77.a.(0.7)500(0.02)tN; at0twe have0(0.7)1. Thus,1500(0.02)10.Nb.50.7500 0.02N0.16807500 0.02259Exercises 0.5 __________________________________________________________________1.2103xxa.The largest exponent is 2. The degree of thepolynomial is 2.b.The coefficient of2xis–1.c.The constant term is 10.d.It is a polynomial of one variablex.2.49527xxa.The largest exponent is 9. The degree of thepolynomial is 9.b.The coefficient of9xis–2.c.The constant term is 7.d.It is a polynomial of one variablex.3.23714x yxy za.The sum of the exponents in each term is 3and 5, respectively. The degree of thepolynomial is 5.b.The coefficient of3xy zis–14.c.The constant term is zero.d.It is a polynomial of several (three)variables:x,y, andz.4.5236275xx yya.The sum of the exponents of each term is5, 5 and 6, respectively. The degree of thepolynomial is 6.b.The coefficient of6yis–5.c.The constant term is 0.d.It is a polynomial of two variables;xandy.5.52235xxa.nna xmeans52a.b.30a(Term is30x)c.23ad.05a , the constant term.6.35417xxa.35ab.14a (Term is–4x)c.20ad.017a7.24xxWhen2,x 2244( 2)( 2)8412.xx   8.222106 4When1,106 4106 4110150140.xxx   

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