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### midterm-sample-4

Course: CS 143, Winter 2011
School: Washington
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Word Count: 1784

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143 CSE Sample Midterm Exam #4 (based on Summer 2009's midterm; thanks to Alyssa Harding) 1. ArrayList Mystery. Consider the following method: public static void mystery4(ArrayList&lt;Integer&gt; list) { for (int i = list.size() - 2; i &gt; 0; i--) { int a = list.get(i); int b = list.get(i + 1); list.set(i, a + b); } System.out.println(list); } Write the output produced by the method when passed each...

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Washington - CS - 143
CSE 143 Section Handout #12Practice Midterm #51. ArrayList Mystery. Consider the following method:public static void mystery5(ArrayList&lt;Integer&gt; list) cfw_for (int i = 0; i &lt; list.size(); i+) cfw_int element = list.get(i);list.remove(i);list.add(0,
Washington - CS - 143
University of WashingtonComputer Science &amp; Engineering 143: Introduction to Programming IICourse Syllabus, Winter 2011Instructorname:email:office:office phone:office hours:Course AdministratorMarty SteppCSE 636(206) 685-2181see course web sit
Ain Shams University - ECON - 101
Chapter6FatigueFailureTheoriesFatigueFailuresssOccurs when stresses are changingthroughout the life of a part._ starts with acrack that propagates until a catastrophicfailure occurs.This usually begins at a manufacturing defector stress concen
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Goldsmiths - COMPUTING - 2910108
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National Taiwan University - EECS - 101
1.1 Answering machine Alarm clock Automatic door Automatic lights ATM Automobile: Engine controller Temperature control ABS Electronic dash Navigation system Automotive tune-up equipment Baggage scanner Bar code scanner Battery charger Cable/DSL Modems an
National Taiwan University - EECS - 101
CHAPTER 22.1 Based upon Table 2.1, a resistivity of 2.6 -cm &lt; 1 m-cm, and aluminum is a conductor. 2.2 Based upon Table 2.1, a resistivity of 1015 -cm &gt; 105 -cm, and silicon dioxide is an insulator. 2.3 I max 2.4 10-8 cm2 7 A = 10 1 = 500 mA (5m)( m) 2 c
National Taiwan University - EECS - 101
CHAPTER 33.1(1019 cm-3 )(1018 cm-3 ) = 0.979V NA ND j = VT ln 2 = (0.025V )ln ni 10 20 cm -62( 11.7 8.854 x10-14 F cm-1 ) 2s 1 1 1 1 w do = + 19 -3 + 18 -3 (0.979V) j = -19 10 cm q NA ND 1.602x10 C 10 cm w do = 3.73 x 10-6 cm = 0.0373m w do 0.0373m w d
National Taiwan University - EECS - 101
CHAPTER 44.1 (a) VG &gt; VTN corresponds to the inversion region (b) VG &lt; VTN corresponds to the accumulation region (c) VG &lt; VTN corresponds to the depletion region 4.2 (a)&quot; ox -14 3.9o 3.9 8.854x10 F / cm F nF C = = = = 6.91x10-8 2 = 69.1 2 -9 Tox Tox 50
National Taiwan University - EECS - 101
CHAPTER 55.1 Base Contact = B n-type Emitter = D 5.2v BC iB + B + E iE C iCCollector Contact = A n-type Collector = FEmitter Contact = C Active Region = EFor VBE &gt; 0 and VBC = 0, IC = F I B or F =IC 275A = = 68.8 4A IBR =0.5 R = =1 1- R 1- 0.5 IC
National Taiwan University - EECS - 101
CHAPTER 66.1(a ) Pavg = 1W 10-5W / gate = 10 W / gate (b) I = = 4 A / gate 105 gates 2.5V6.2(a) Pavg = 100 5x10-6W / gate = 5 W / gate (b) I = = 2 A/ gate 2.5V 2x10 7 gates(c) I total = 2(2x10 gates)= 40 A gate7A6.3 2.5 - 0 5 (a ) VH = 2.5 V | V
National Taiwan University - EECS - 101
CHAPTER 77.1' n -14 cm 2 (3.9) 8.854x10 F / cm 3.9o K = nC = n = n = 500 Tox Tox V - sec 10x10-9 m( 100cm / m) &quot; oxox()F A A = 173 x 10-6 2 = 173 2 V - sec V V p ' 200 A A &quot; K 'p = pCox = Kn = 173 2 = 69.1 2 n V V 500 ' Kn = 173x10-67.2VDD(5 V)
National Taiwan University - EECS - 101
CHAPTER 88.1(a) 256Mb = 28 210 210 = 268,435,456 bits (b) 1Gb = 210 = 1,073,741,824 bits8 10 10 28( )( ) (c) 256Mb = 2 (2 )(2 )= 2I pA 1mA = 3.73 28 bit 2 bits( )3| 128kb = 2 7 210 = 217 |( )228 = 211 = 2048 blocks 17 28.28.3(a) P = CV (b) P
National Taiwan University - EECS - 101
CHAPTER 99.1 Since VREF = -1.25V , and v I = -1.6V , Q1 is off and Q2 is conducting.vC1 = 0 V and vC 2 = - F I EE RC -I EE RC = -(2mA)(350) = -0.700 V9.2 V IC 2 0.995 F I EE = exp BE VBE = 0.025ln = 0.132V IC1 0.005 F I EE VT (a) v I = VREF + VBE = -1
National Taiwan University - EECS - 101
CHAPTER 1010.1 A/C temperature Automobile coolant temperature gasoline level oil pressure sound intensity inside temperature Battery charge level Battery voltage Fluid level Computer display hue contrast brightness Electrical variables voltage amplitude
National Taiwan University - EECS - 101
CHAPTER 1111.1v O = vS iS = v 1M 1k (1000)1k + 0.5 | Av = vO = 990 or 59.9 dB 1M + 5k S | Ai = iO 990 6 = 10 = 9.9x105 or 120 dB iS 1000 vO 5V = = 5.05 mV AV 990 vS 990vS and iO = 1M + 5k 1kAP = Av Ai = 990 9.9x105 = 9.8x108 or 89.9 dB | v S =()11.2
National Taiwan University - EECS - 101
CHAPTER 1212.1(a) A = 10 20 = 2.00x104 | Av-ideal = 1+A Av = = 1+ A FGE =86150k = 13.5 12k2.00x10 4 = 13.49 4 12k 1+ 2.00x10 162k 1 13.5 -13.49 = 6.75x10-4 or 0.0675% | Note : FGE = 6.75x10-4 A 13.5 2.00x10 4 = 125 1.2k 4 1+ 2.00x10 151.2k 150k (b
National Taiwan University - EECS - 101
CHAPTER 1313.1 Assuming linear operation : vBE = 0.700 + 0.005sin 2000t V 5mV vce = (-1.65V ) sin 2000t = -1.03sin 2000t V 8mV vCE = 5.00 -1.03sin 2000t V ; 10 - 3300IC 0.700 IC 2.82 mA 13.2 Assuming linear region operation : vGS = 3.50 + 0.25sin 2000t V
National Taiwan University - EECS - 101
CHAPTER 1414.1 (a) Common-collector Amplifier (npn) (emitter-follower)RIQ1viR1R2+RER3vo-(b) Not a useful circuit because the signal is injected into the drain of the transistor.RI viRDM1+R3vo-R1(c) Common-emitter Amplifier (pnp)
National Taiwan University - EECS - 101
CHAPTER 1515.1(a) IC= F IE =VCE = VC - (-0.7V ) = 5.87V | Q - Point = (20.7A, 5.87V )1 F 12 - VBE 1 100 12 - 0.7 = = 20.7 A | VC = 12 - 3.3x105 IC = 5.17V 5 2 F + 1 REE 2 101 2.7x10 (b) Add= -g m RC = -40(20.7A)(330k)= -273Rid = 2r = 2 oVTIC=
National Taiwan University - EECS - 101
CHAPTER 1616.1 Av (s) = 50 s2 s2 | Amid = 50 | FL (s)= | Poles : - 2,-30 | Zeros : 0,0 (s + 2)(s + 30) (s + 2)(s + 30) s rad | L 30 s (s + 30) | fL = Yes, s = -30 | Av (s) 50 fL = L 30 = 4.77 Hz 2 22 2 1 302 + 22 - 2(0) - 2(0) = 4.79 Hz 2 50 2 | MATLAB
National Taiwan University - EECS - 101
CHAPTER 1717.1(a) T = A = (b) A = 10Av =80 20|Av =1=5|FGE = 0= 10000 | T = 10000(0.2)= 2000A 10000 100% 100% = = 5.00 | FGE = = = 0.05% 1+ A 1+ 2000 1+ A 2001 A 10 100% (c) T = 10(0.2)= 2 | Av = 1+ A = 1+ 2 = 3.33 | FGE = 1+ 2 = 33.3% 17.2 1k
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J(4uetq 6n I,_._Head- )gFt: ?-barnWq*erflow =SOFL3 :o.r.1*39. = boolo'l) 0:9rLq4: 9'8l xO.bxo.tg*3r?,6&gt;r&quot;,:6,:)', ,:i-,f * (&quot;g,.-grevrlgforce- q-&amp;th/satL= etKqe\$cgO - q,ucnti+y oF urqtercfw_+ _- e*Fe&lt;_cfw_vL hecrd en )Tr) 6,3rLOOiir
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.lqFF+JOqeEcfw_rbn 2-,i) Per = Pgh = rooo Vgl^t x g&quot;slV*exgxPerrn/s. x t00rv1Hloog &amp;,urt* : VXtoOO l.g/n3x 1.gl rn/s.lrt00V: lolg,3?r.n3qre _ tot q .3T t,)/ uonn= tot.931*.,j, ThiS hy/ro is*oiirecrsonqbtegroc,luce \ooo-: :.,=,o '
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Keller Graduate School of Management - HR - hr595
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