Homework_Solutions___Week_1

Homework_Solutions___Week_1 - 3mm II" “II"...

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Unformatted text preview: 3mm: II" “II" I—fllflrfl'flg Jfllllll 01' Solutions for Exam l—Morning Section t N 1. E: ere {Lg-1 = $34.3 Em lumen = (132%) mg} _ 111130.01] _ t— {4.3 dsys}( “(fl-5i ) — 28.6 days Answer is C. 2. phi: least 2}=p{2 orfloI-ior 5} = l — p-[fl or 1} = 1 - Fifi} -P'[T} Use binomial distribution to find the individual prebs— bibties. p = Isl-sibebilitg.r of heads 2 13.5 e = probability of tails = [1.5 p{e} = {43.515 = 0.03125 p-{l} = (fifssflesr = {3.15325 p-[st meat 2} 2 1 4.03125 # sissss = seiss Answer is C. 3- {1N2} - {4H3} = -10 Answer is A. 4. {2“—e$+s]+{y2—4y+~1}—12=e+4 {3‘3}2+(s~2}2=95 r=~.fl'r2_5=5 Anselm-15C. 5. Solve by featuring. 2— _ 11mE g=1im —{I+3H$ 3} 2—6 3—3 :t—I3 3‘3 =lfli¢+flj :6 H Answer is C. .l-‘ll. —-.-.-.t- E. The magnitude of vector V i5 v = [139+ (231+ {111 = 9’5 The :r—directien cosine is i V”: 1 cos =~—=— at V E 1 iizees 1" (—) =659° fi AnswarisC W1 in" —1 --l T f—2d$=—— =—--~— #1}? Answer is C. 1+4+?_ 3 d E. mean = The ssemisxd deviation is in h a)“ + [ct—4}2 + (L4) 1 a=¥+=ufi =2.45 Notice that 11 ~— 1 would have been used if the sample standard deviation hsud been requested. Answer is A. Q. The magnitudes of the ten: teeters are V1=J|11J2 +12}2 +{1}9 = V'E is = {112+ {3F +i—1’F = v’E 6 = $5-: [use + {23:33 + {Iii—U] = w NE} {fl Answer is E. '.| 11]. A=f e’ucie=eJ =e]*eflI s 1 fl 5:: 1.715 Answer is B- — Preisuinnni Plliliiillijflm, ITIE. ._.-._-.-_-.-- ___ _.___...__:. ._.__: ._. _...-..'.-._"_'.3:-.¢4L.'LLS.'_T.L".LL.£p_-quq.mmmm_._u__. In HIE"- SIIIPIE Eflfl’lilll‘ifli‘ls w 1'1. Ths first and second derivatives are 15. Differentiating. g=£3_3¢+2 y*_—.2?m2+2m—15 yii=3$2-3 y"=54.'r+2=fl u.- = L; = ~U.fl3i 34 For a critical point, y’ =‘ D. B]: inspection {bsseti DI! the four answer aim-ices], y“ = i] at :r = I and :r = -1 Elli—D33“ = 32-35 { {-0.U3?,32.5tij y”{1}= (3)131}? — 3 = n iii—2:- = {SH—22F — 3 : 9 _ “‘23 E ii} {'Ei’i _ (LEE—2}: + {EH—23' + 5 5 ‘1 16. HR! = mil—+2 — 1]2 + [3 + T + 4}“ + {4 — 1 + 2}2 = Miss + 25 ' = 22:: = 15 Answer is C. Answer is A. 12. 3:1: ‘1 l —1; ‘1 u“ =fiU-25D d-fl {a} 3-H E E) Answer is {3. Answer is C. 1?. Multiplying through by 2 gives 13- {y=+y+ii+{x2—2x-r-1i=5+-f;+1 :r”+Sm’+1fir:1CI = 253% The characteristic equation is 2 2 .. y+l +{z—1)=2ay4 ( 2} r2 + sr + is = s This is ECETCIE 331111313351 3+- ili —11'r2i|1 5'3 E'mu is at 3 '2 1' The mots of the Characteristic Equatifln 9.11:: Answer is C. The homogeneous {natural} Immense is EmLEuraJ = 443-41: + Big—4e By inspection, 3 = 51,13 is a. particular sslution that. subsets the nanhumngunsous aquatic-n, so thc tats! IESPGIIEE is z =As—4'z +His'41+% 5511432121 sti=fl, .- i“ i l—AL +3 1 _ E Differentiating r. m“ = [§} {—4}e‘4* + 131— 415’“ +e"“‘] + Er Sinscr‘zflstt:fl, 1| —% +B[U+1] I1 ETEHI in}? + %s‘{11 In)2 + EIWEE 111]2 = 4934,1111: 4', U B :r: Ea": + git—J” + i Answer is. D. Answer is D. Inn-F. ‘2...» Prnfessimmi FUHiEfljim‘i, In[, M _ mthmI—flflflqm N 13. Since Inghhfl’] = Hm], Agavmr is C. .. _ 11111an of orange balls _ 7' 19‘ Whammy _ tutal number nf b.1115 _ 1—7 = 114118 Answer is D. 3D. Since 3;" = 1-1, this is a. maximum value. yj=14Jn—H FT] :1: =an4 24%) = (ET — (a) +3 = m (%) - {'14 + a =215f23 Amer Is D. 21. From the double angle formulas. sin ‘25 = 2511130059 Answer is A. 22. Use Euler’s furmuLl. :9 = cmfl+i3infi H=1Tfh2 23. 3'5. graph of the area. is aha-ML Since 3; = V's! — 1:! is the aqmtinn fur the top half of a. circle. the volume after a revolution wifl. be a hemi- sphere. V = fiVnpm = G} {Elma = H} {EHEEF = lfi.?fi Answer is C. 24. Guns products can only be performefi with vectors. hum: is D. m. Int. In To: Mike Timmons [email protected] Subject: Hypothesis testing problem Hypothesis testing problem An engineering wishes to determine if a chemical analysis provided by ABC Laboratory is biased. Ten standard samples with a known concentration of 25 mgl'l were sent to the laboratory at different times over a 2 week period. The laboratory return estimated concentrations for each of the“ ten samples. The average of the ten values was 25.61. whereas the sample standard deviation of the ten values was [21.86. a) Using this data what is a 95 percent confidence interval for the true mean concentration of the values the ABC Laboratory will report when sent standard samples with a mean of 25 mgl'l. b} Using a test with a type Ierror of just 113%. would you conclude with this data that the mean of laboratory analyses is different from the true value of 25 rngl'l for the standard solutions? SDLUTiDN n to average 25.61 stdev 0.86 a} 95% confidenoeinterval df 9 alphai'2 2.50% tevalue 2.262 upper 26.23 lower 24.39 bl Hypothesis test t 2.2430 critical value for a two-tailed test with 5% in each tails is 1.333 Here the value of t = 2.24 is much greater than 1.833, so we reject the hypothesis that the mean of the laboratory results is 25. K ...
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