1105_LW5B_solutions_F11

1105_LW5B_solutions_F11 - 2 each part below y is an...

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Unformatted text preview: ' ' 2.. each part below, y_ is an exponentiai fimcfiqn ofx. You must mite the formuia. . I I I a) in a laboxatmy setting, a population of flies starts at 260 flies and triples every W961; :Wfite 2m exponentiai fuflctionformula that gives 52,1113 pop‘giation 3: weeks from'now. - AgsWer:;y_=.z§° (3} b) Quantity-y starts at milligrams and i5 halved each day. ;. Wztitean exponergfial {311191ng __ formuia that gives {he y—quantiiy 3 days from ndw. ' .. .- : . - . . '1 ' . . X -_ Answer: 3/: Zaopéj c) -. Quantity starts at 2.139111131331115 grows by a factor of 11165 _ _ 'Write'an exponentiai funcfidnfiomula fihat gives the y-quantity _x___h_ours_ now,_ ‘ -. d) :Quafitity y starts.at100‘mflfigrétms decays by a' faciw of £1,558 every hofir,‘ ' ' . 'Write apexponemiai funggion formula that givesi'the y—qfiailtityixihoms from HOW. . - ' - - _ g 7. “a _. -- Answer: 32: Z“§_{D’5$322; . 3. For each exponential function formula below, détcrming whether thé: fufiiéfiién rafiréséms; éfiponential growth or decafithe initialqilanfity, the growth. or :1er factor, and the grow}: or decay rate. Sketch its graph, labeling the "vertical intercept. Your sketch should ahow the asymptotic behavior bf ' the graph} and two other points, 9113 with positive ipput and the otherflith nggafive input. _. -= 2 _ '- ' '- _ E 73:) Q(z)=(1.2x10‘?)(1.17)' L b) P(¥}_;30(0.85)‘ ' - ' '6.) Twp-25(4)"; _ resemLPp-fi - I I - - .Jecé. ' g; I “<3 m . q‘hflfaj rajue, “rt. I:ZK{0%'-'~ r - '3‘“; ijéwa'22305 '- I I )I'fl I ' - f“ magma, 1 ' é; {Fm‘zévfimegfj a; P «L I I raggawfiwswigv, -5!” ' ' 3 .. - 4. The four-graphs shown correspond to the four function foxmuiasbelow, Mags}: each graph with its fogmuie by filacing..the lefterpf the graphite the next to its formula. ‘94 / 'grnwfifdeee;.- W!“ . -- - _J 3. ' ; -¢- 53 32:00.1)? [teem e?» {oxalgrmfih _ ;_mee;e=- yecwm*.+43© e::¢%ere ‘ ' 2_ . . yéCQY; 333 +héo ..§ao‘2, 0:9 I' I 5e! ,a..._—._._._._-...._., y=c£078>x 4 an 2:3» 2,32 am: a w 5. When first taken, 2} cold medicafiogl intreducee 200mg of e active dreg into the bleed _ The I. . I i r ametmt of ecfive drug (ieceys exponentieiiy aeeercfing to __the formula .50) :‘200(0.7_Q7)_’ _:;_ -' ' - a) Use the calculator’s table and graph features to estimate how long it will take for there te-be only - - . - 150 mg ofaciive drug inthe bloodstream, rounded 19 the nearest tenth of anhour day. - ' ' x:memw~-_e 55. 1935 , - - - ' - ' ._'A .%. -‘ werztss_-a_-'_ours .- Y22'15ifi) I ZIgCALqfinZEfSEC—i— I . _ _ . 13) Use the caieulator’s tableand graph feauues teeetimate thd half Iifefer. this. drug, rounded to the _- . We tethofehw- -* J ' " - - : . ' ' - .-.+:memm «Fee m we yi :190. (6.7197) .' '5 '.'--An_swer: f2: 2,0 . hours be {flosng-gu 19075913 Yea-1G0 ' 1“";ce\c,_1n reefer .'x:;.qfi912.3¢1m -. . _; " .- - -. -. . - _ - . .- ) Use the realcuiator’s table and graph features to {asexuate when only{§5%roffheinitia1amounbjs - _ still a five inthe blood stream, mended to the neaxest tenth of an hem. _ 3- ' '61_§'f){;¢g3§ 2;; 3 Q .¥ #3:! .Answex:1re bib hours 7-" '- 25", caiqiniéme : Keiwetre‘: r - ' .- ' - .. - “- (1) Sketch the glaph andlabel the points 01:; the graph conesp ding to you: answers from a)-c). _ _ _. it; beef-0.37? ' Yam-ca_ns+aei°- .150! emmgfeeud 3g) 6. A muiual fund account is currently velued at $20_0.0._ _ a) .Ifthe account damages f9 $2200 eve; seme time period? What is the piiércefit: incmage Dim. 'thistimepefiod? " -' '- - ' . . . . Em? ' Beale-L '. - . . . '” ' ' anewmee ‘ ; ’63-‘35 --Z - j if}: 1&0?“ e_ :10 W. I b) Ifthe aceeuntde'ceeases to $1700 osfe‘I-flfiefimeperiod,whet_'isthe15erce1}t decreaeeover. U thistimeperidd?-- " -. .. .. _. ._ _ . ._ : .moewaeaa ' MLwWMM 2x we in »_ . we) .—~ . 5 . .. lb. " . gems ;--; ;- . _: -_ %_ :5 7. The vaiue of'a condog'ninimn is currently $100,000. ' -' . If its value Epcreascs by 10% pic: sometime periodgfmdflge increase in vahw'andjhe _- .3 ' condo_"s_ya1ue afihe 3nd pf.t_hi.fs_time__peri0_d._ .-; ' - - - ' .- ' iogégogiff’ AIQGO‘W‘Q‘C’) {.Oféjgxgwers:§aqea§einvahie:$ $0 {363$}; :11 ' __eahc\_\rcg\¢e_ 353190,??? % \ogooéfiimflmQ Césido’é.valua=-.$Ji_ci£%€3_f' ' Now We I b) In part 20,133! what factor did the céfido’s' value increase? That is, ifwéfimitip.1y$100,000 by _ i_ _' the number rIg lb. b.“ we get thq condqfs Ivalu'e' at the and ofithis Lime period-s __ ' ’ ; . ' . . . c) If its vaiue decréasés by 15%,dersome tinié period, find #13 dBCIéaSBIiI-l valus and the _ . -' 'condo’s vakue at the and ofthis time periqd. - ' - ~ -: - . . . 15 c? :39 5-100 000 is 4‘; £09 000) i , . _ . I .c: 1 _ f _. - . Answers;I‘Decmasein-Value:$ 1500C) Qfla “‘91.”?ng 55:36:00? ggggwg- _ -. " '- .' -- - = - . Mob: 2625??? - swfi'fiéfl 45-353 fi-iwgééfiifii‘é? 7' d) In paIt‘a), by Whatfimmr did the condo’s Value decrease? Thai is, if we mlfltipiy $100,090by ' ' the number 5 .5 w; gst tbs condok value at the and ofthis tinge pgrioti. _ ' - 3 ' ' r m___ ‘8. Thepopulation of a city grows annually by_1_._’;'%. 3' ' a) If thé population is 40,500 at some particuiaxpojjfi in time, findthe repulation one ygér’lat_ei._ 405‘00 {infill-dadiéimenmwhgicnumyqr)I l 8 I b) In general, if wemultiply the popfiiation at fifiynfibmefit by the numbef s '0' 2'7? , get I ‘ '. thepopuiationone yearlatgrg '- ' ' ' -- ' " r .. ' -' -‘ c) Ifthe current popuiatidn is denotégi , mite a fiméfiog formfila lhéfixépréséntsfibpuiati'ont .. yams into the future. Use this fimcéfion anda graphic approach to detennizfile howman: ears _ Will‘ka o ulzition t9 613;}ng £9.35 hkg' F9 '3er +6 :2' ' “2 - 4»; '. ' ' " ' I _ :4._;_ . ' 3:, . . WV, .--.—.-=w- ?f. as. . - .. v‘ 1.3. 6:: (2,92?) ' x . (qu‘m c; jig: \{f I Calculefinngegound Interest « "Finife Gama-011mlng . If 1? 2 original invesiment (called princépal), r : ennuai (call-ed nominal) interest rate in decimal form, 7 1'2 =_.t11_.e_ number of times each year_i_1_1ter_esti_s cqmpounded, and t: the number of years over’which interesprate is compolmded, then account value A is given by A; Pb +£J . The annual interest rate . ,' is efien called the nominal rate. When'interest is compounded moreofien than once each year (that is, '5 . I; > 1),__acco_unt vaiue actualiy grows faster than the nome rate suggests. faster Iate__is (failed the , Effective meme —:1]x‘10_0 gives this effective rate as epegeentage, . ' .9. An infiestment of $4000 eanis 7.2%annnal (nominai) interest Round money to the nearest penny, - I . I. _ and effective interest rates t_o the neaxest thousandth of e'pereent __ - - . - - -- . . ._ : _l___d_fi__ r- “Tit-1+ . I - ;;_ ' -. ' ' :- . I. ._ a) If iniexest is @mpoundtmwtfl the account value after 10 years andjhe efi'eefivexate. _ “Acesgiw50164r “View???” ' ' - Ew'gg‘feflx we '2. 713%7932‘186“ I I - Haifa-rm A<sé).-e 40990.93)‘L‘Qég‘ggugég -'b_) If imam is compounded daily, find ihe account value after :0 yea-rs and the efl’eCfiveiete. - I _ ,372,\3{o§+ --'-3,;,g' =~- ' 7' , MBMWW“ 3e) i 3[HHEET.‘el1Xieee7q%¥W-?3355 I I I ' " I I ' ml -_ ' ' . . - I A02) 4 ' 3635.013”§§§@fi“ ’ - , ,, r a fire-~90? we??? ,ge‘fiflwlie - NaturaIEx onenfial Function Continuous Growthffiece and Continuum Com eufidin As n —>¥_oa , file quanfity [I ~> e 231828182846 5 an irratiefial fififilber' called Euler’s number. ' 7 I The co‘nstante is the base for the natural exponential function f e e_’_‘ . ' when. the annual (30mm) . rate is cempeeuded copfinuously, account valfieA given by I Peri},- [54:1]me flm : effective rate as a percentage. 30. An invesflfient of $4000 earns 7.2% meal (momma!) interest. Find the-aeeeufit value afiefi 10 years 3 __ . and the effective rate. Round moneytethe nearest p_enny,-and_the effeeiiye__inte1jest rate to the ' ' nearest thousmdmpf a perceet.-” '_ _ - . .' '. ._} r " ' " ' .Fnri w ';.b’?l- 1'11 ' ' ' C- - -_- ,,. mo .3: U geiwgfifl W I I H I flJw’v-‘ihsk; {fjflkp‘ffg.1.p—rrsij:ffi . . 3:4. .-: 1.17:”: me? e 11. The amount of drug stifl active inihe biood. stream thaws after the._dmg was first administered can ' sometimes be medeled by a continuous decay model. Suppose the concentration of a ' ' pmfienlar drug is 2 mifligrams per liter of biood, and conceptration decays at aeominuous rate of 2%. . - " ' pezheur. Find a fommla ofthe fem A (2) =- Aoe“ that gives the coneeetx'atien i: house-fie: it _ u . ' has been admimstexefi, and use a graphic app each to estimate how-long it wili take for ' ' cezlcentrafioetoreach 1.5 mg/L. ' - " =_ ' - ' '3' ' II I ' ' "3 _ - '-. 15 5152's: 02" ' '_ se‘szq..wggsg . 12. The accident on April 26, 1986, at the nuclear power plant in Chemebyl, Russia, is censidezfefd .the worst nueiear disaster of the nueleazr age. Large amounts 9f the radioactive Substance CesiiuliszIS’l _ Welfe emitted into the air. ( hfipzifegjdkipedie.orgiwiki/Chemobyi 'di'saster ) The amount of an ' initial 1 (JO-milligmm sample of Cesium—137 remaining'afierx years isgiyen by'the centijmoes.deeay . model $10094.022951 . ' . I ._ . ' . . . la) ‘What continuous .yeaxiy decaygate does the formula suggest? - fa ‘ Z?! 5 pegyear. ' b) The half-life of a decaying substanee isihe amount oftimefer the'sfibstance {e :deeay by I. :5' _ 50%. Use a graphic approach to estimate the halfrlife 0f_Cesium-137,_ rounded to thenearest ;_- - year. Sketch the graph that iliustxexies your strategy. 'g woflzZQS'x. Half—life Is 53:15 ‘ 'r"__'yea_rs_ ' .. 50:21:30 3 ' ' " kg; If 6’ mo Wei}? 9)) e) _ natural exponential model ofthe form A (x) '; Aces-iv _.¢gm_be Iem‘itien the fame” '_ 3 I _ gm z w by letting c 2 A0 and sf; (2. Fem base a fog m) : msg#29323; made to I . four decimal pieces, anduse this reuhfie'd base to rewrite the nafeiel expenergfigl Inc-fie} for ' Cesium-13? i11_'the_:fonn A(x)= Cra‘1 3._ ' " " ' ' ' ' ' ' " -_ F- x 5&02235x ' I _ .. < _ I {Iii-K mafia ' J ' _. ' AcessA ($27-33? _ . - l' I _ 0 . 4;) rm“; .022”; - CQeclsLs$3en3 thriiiwms. qnmgg‘égee 7 Ne 9:? 1*?” ‘3le” '- -- pad-p, 2,2613% 55 aifimiéfi‘fk doom? _ ' - r gggj%§1rxg$ emer ésgfigrf {gig-‘3 ...
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1105_LW5B_solutions_F11 - 2 each part below y is an...

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