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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 {Lg1 = $34.3 Em lumen = (132%) mg} _ 111130.01] _
t— {4.3 dsys}( “(ﬂ5i ) — 28.6 days
Answer is C.
2. phi: least 2}=p{2 orﬂoIior 5}
= l — p[ﬂ or 1}
= 1  Fiﬁ} P'[T} Use binomial distribution to find the individual prebs—
bibties. p = Islsibebilitg.r of heads 2 13.5
e = probability of tails = [1.5
p{e} = {43.515 = 0.03125 p{l} = (ﬁfssﬂesr = {3.15325
p[st meat 2} 2 1 4.03125 # sissss = seiss Answer is C. 3 {1N2}  {4H3} = 10 Answer is A. 4. {2“—e$+s]+{y2—4y+~1}—12=e+4
{3‘3}2+(s~2}2=95 r=~.ﬂ'r2_5=5
Anselm15C.
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=J11J2 +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$23 y"=54.'r+2=ﬂ
u. = L; = ~U.ﬂ3i
34 For a critical point, y’ =‘ D. B]: inspection {bsseti DI! the
four answer aimices], y“ = i] at :r = I and :r = 1 Elli—D33“ = 3235 { {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“ =ﬁU25D
dﬂ {a} 3H E E) Answer is {3. Answer is C.
1?. Multiplying through by 2 gives
13 {y=+y+ii+{x2—2xr1i=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'r2i1 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 = 44341: + Big—4e By inspection, 3 = 51,13 is a. particular sslution that. subsets
the nanhumngunsous aquaticn, 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.
InnF. ‘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;" = 11, 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 ahaML 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 twotailed 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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 Spring '07
 TIMMONS/ANESHANSLEY/

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