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Crib Sheet 3

Course: BUS 360, Fall 2007
School: Wisc Stevens Point
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path Critical is LONGEST path from start to finish. There is no slack in the critical path. There IS slack every in path not critical. Gantt chart (bars) doesn't show precedent relationships. Pert + CPM do.

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Coursehero >> Wisconsin >> Wisc Stevens Point >> BUS 360

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Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
path Critical is LONGEST path from start to finish. There is no slack in the critical path. There IS slack every in path not critical. Gantt chart (bars) doesn't show precedent relationships. Pert + CPM do.
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Wisc Stevens Point - BUS - 360
Wisc Stevens Point - BUS - 360
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Wisc Stevens Point - BUS - 360
HOPEThe ThailandNo birth certificate. No future. No hope. Until now. The Thailand ProjectProjectGet the Facts. Get Angry. See Red.The Thailand Projectthe programs-Higher Education as Humanitarian Aid: Provides a college education to studen
Iowa State - LSCM - 360
LSCM 360 Business LogisticsWelcome. Good Luck. Learn Lots. Study Hard. Be Scared.SUPPLY CHAIN MANAGEMENT DEFINEDSupply Chain Management (SCM) encompasses the planning and management of all activities involved in sourcing and procurement, conversi
Iowa State - MGMT - 370
Chapter 7 Management 1. Decision making and planning process a. The planning process takes place within an environmental context. Managers must develop a complete and thorough understanding of this context to determine the organization's mission and
Iowa State - MGMT - 370
Chapter 8 1. The Nature of Strategic Management i. Strategy- A comprehensive plan for accomplishing an organization's goals ii. Effective Strategies- A strategy that promotes a superior alignment between the organization and its environment and the a
Iowa State - MGMT - 370
Chapter 9 1. The nature of decision making a. Decision making defined i. Decision Making- The act of choosing one alternative from among a set of alternatives ii. Decision making steps 1. Recognizing and defining the decision situation 2. Indentifyin
Iowa State - MGMT - 370
Chapter 12 1. The nature of Organization Design a. Organization design- The overall set of structural elements and the relationships among those elements used to manage the total organization. 2. Universal Perspectives on Organization Design a. Burea
Iowa State - MGMT - 370
Chapter 13 1. The nature of organization change i. Organization change- Any substantive modification to some part of the organization b. Changing Technology and operations i. Enterprise resource planning- A large-scale information system for integrat
Texas A&M - PHYS - 218
1.4:g 11.3 cm31 kg 1000 g100 cm 1m31.13 10 4kg . m3
Texas A&M - PHYS - 218
1.5:327 in 32.54 cm in31 L 1000 cm 35.36 L.
Texas A&M - PHYS - 218
1.6:1 m31000 L 1 m3 2111.9 bottles1 gal 128 oz. 3.788 L 1 gal 2112 bottles1 bottle . 16 oz.The daily consumption must then be 1 yr bottles 2.11 103 yr 365.24 da5.78bottles . da
Texas A&M - PHYS - 218
1.7:1450 mi hr 1.61 km mi 2330 km hr . 2330 km hr 10 3 m km 1 hr 3600 s 648 m s.
Texas A&M - PHYS - 218
1.8:180,000furlongs fortnight1 mile 8 furlongs1 fortnight 14 day1 day 24 h67mi . h
Texas A&M - PHYS - 218
1.9:15.0km L1 mi 1.609 km3.788 L 1 gal35.3mi . gal
Texas A&M - PHYS - 218
1.10: a) 60mi hr ft s21h 3600 s 30.48 cm 1ft5280 ft 1 mi 1m 100 cm88ft s m s2b) 329.8g c) 1.0 3 cm100 cm 1m31 kg 1000 g103kg m3
University of Texas - ME - 330
University of Texas - ME - 330
University of Texas - ME - 330
Texas A&M - CVEN - 221
Texas A&M - CVEN - 221
Texas A&M - CVEN - 221
Texas A&M - CVEN - 221
Texas A&M - CVEN - 221
Texas A&M - CVEN - 221
Texas A&M - CVEN - 221
Texas A&M - CVEN - 221
Texas A&M - CVEN - 221
Texas A&M - CVEN - 221
Texas A&M - CVEN - 221
Texas A&M - CVEN - 221
Texas A&M - CVEN - 221
Texas A&M - CVEN - 221
Texas A&M - ENGR - 221
TEXAS A&M UNIVERSITY ENGR221/502 Introduction to Engineering Mechanics Fall 2005Final ExamDecember 9, 3-5 pm, 2005Closed books and closed notes. Needed formula sheets are provided in the exam. This exam has 5 questions for a total of 100 points.
Texas A&M - PHYS - 208
Chapter 2CHAPTER 2 - Describing Motion: Kinematics in One Dimension 1. We find the time from average speed = d/t; 15 km/h = (75 km)/t , which givest = 5.0 h. 88 km/h.2. 3.We find the average speed from average speed = d/t = (280 km)/(3.2 h) =
Texas A&M - PHYS - 208
Chapter 3CHAPTER 3 - Kinematics in Two Dimensions; Vectors 1. We choose the west and south coordinate system shown. D 2x For the components of the resultant we have W RW = D1 + D2 cos 45 D 2y D 2 = (200 km) + (80 km) cos 45 = 257 km; RS = 0 + D2 =
Texas A&M - PHYS - 208
Chapter 41CHAPTER 4 - Dynamics: Newton's Laws of Motion 1. We convert the units: # lb = (0.25 lb)(4.45 N/lb) ~ 1 N. If we select the bike and rider as the object, we apply Newton's second law to find the mass: ?F = ma; 255 N = m(2.20 m/s2), which
Texas A&M - PHYS - 208
Chapter 5CHAPTER 5 - Further Applications of Newton's Laws 1. The friction is kinetic, so Ffr = kFN. With constant velocity, the acceleration is zero. Using the force diagram for the crate, we can write ?F = ma: x-component: F kFN = 0; y-component
Texas A&M - PHYS - 208
Chapter 6CHAPTER 6 - Gravitation and Newton's Synthesis 1. Because the spacecraft is 2 Earth radii above the surface, it is 3 Earth radii from the center. The gravitational force on the spacecraft is F = GMEM/r2 = (6.67 1011 N m2/kg2)(5.98 1024 kg
Texas A&M - PHYS - 208
Chapter 7CHAPTER 7 - Work and Energy 1. 2. 3. The displacement is in the direction of the gravitational force, thus W = Fh cos 0 = mgh = (250 kg)(9.80 m/s2)(2.80 m) = 6.86 103 J. The displacement is opposite to the direction of the retarding force,
Texas A&M - PHYS - 208
Chapter 8CHAPTER 8 - Conservation of Energy 1. The potential energy of the spring is zero when the spring is not compressed (x= 0). For the stored potential energy, we have U = ! kxf2 0; 35.0 J = ! (82.0 N/m)xf2 0, which gives xf = 2. 3. 4. For t
Texas A&M - PHYS - 208
Chapter 9CHAPTER 9 - Linear Momentum and Collisions 1. We find the force on the expelled gases from F = ?p/?t = (?m/?t)v = (1200 kg/s)(50,000 m/s) = 6.0 107 N. An equal, but opposite, force will be exerted on the rocket: 6.0 107 N, up. For the mome
Texas A&M - PHYS - 208
Chapter 10 1CHAPTER 10 - Rotational Motion About a Fixed Axis 1. (a) (b) (c) (d) (e) 30 = (30)( rad/180) = p/6 rad = 0.524 rad; 57 = (57)(p rad/180) = 19p/60 = 0.995 rad; 90 = (90)(p rad/180) = p/2 = 1.571 rad; 360 = (360)(p rad/180) = 2p = 6.283 r
Texas A&M - PHYS - 208
Chapter 11CHAPTER 11 General Rotation 1.z (a) For the magnitudes of the vector products we have i i = i i sin 0 = 0; k j j = j j sin 0 = 0; j k k = k k sin 0 = 0. y (b) For the magnitudes of the vector products we have i j = i j sin 90 = (1)(1)(1
Texas A&M - PHYS - 208
Chapter 12CHAPTER 12 Static Equilibrium; Elasticity and Fracture 1. From the force diagram for the sapling we can write ?Fx = F1 F2 sin 20 F3 cos = 0; 380 N (255 N) sin 20 F3 cos = 0, or F3 cos = 293 N. ?Fy = F2 cos 20 F3 sin = 0; F3 sin
Texas A&M - MEEN - 221
Texas A&M - MEEN - 221
Texas A&M - MEEN - 221
Texas A&M - MEEN - 221
Texas A&M - MEEN - 221
Texas A&M - CHEN - 323
PROBLEM 1.1 KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermalconductivity k and inner temperature, T1. FIND: The outer temperature of the wall, T2. SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-dire
Texas A&M - CHEN - 323
PROBLEM 1.41KNOWN: Hot plate-type wafer thermal processing tool based upon heat transfer modes by conduction through gas within the gap and by radiation exchange across gap. FIND: (a) Radiative and conduction heat fluxes across gap for specified hot
Texas A&M - ELEN - 214
Circuit Variables1Assessment ProblemsAP 1.1 To solve this problem we use a product of ratios to change units from dollars/year to dollars/millisecond. We begin by expressing $10 billion in scientific notation: $100 billion = $100 109 Now we det
Texas A&M - ELEN - 214
Circuit Elements2Assessment ProblemsAP 2.1[a] To find vg write a KVL equation clockwise around the left loop, starting below the dependent source: ib ib so vg = + vg = 0 4 4 To find ib write a KCL equation at the upper right node. Sum the curr
Texas A&M - ELEN - 214
Simple Resistive Circuits3Assessment ProblemsAP 3.1Start from the right hand side of the circuit and make series and parallel combinations of the resistors until one equivalent resistor remains. Begin by combining the 6 resistor and the 10 r
Texas A&M - ELEN - 214
4 Techniques of Circuit AnalysisAssessment ProblemsAP 4.1 [a] Redraw the circuit, labeling the reference node and the two node voltages:The two node voltage equations are v1 v1 - v2 v1 + + = 0 -15 + 60 15 5 v2 v2 - v1 5+ + = 0 2 5 Place these equ
Texas A&M - ELEN - 214
The Operational Amplifier5Assessment ProblemsAP 5.1 [a] This is an inverting amplifier, so vo = (-Rf /Ri )vs = (-80/16)vs , vs ( V) 0.4 2.0 so vo = -5vs3.5 -0.6 -1.6 -2.4vo ( V) -2.0 -10.0 -15.0 3.0 8.0 10.0 Two of the vs values, 3.5 V and -
Texas A&M - ELEN - 214
6 Inductance, Capacitance, and MutualInductanceAssessment ProblemsAP 6.1 [a] ig = 8e-300t - 8e-1200t A v=L dig = -9.6e-300t + 38.4e-1200t V, dt 38.4e-1200t = 9.6e-300t t > 0+v(0+ ) = -9.6 + 38.4 = 28.8 V [b] v = 0 when or t = (ln 4)/900 = 1.54 m
Texas A&M - ELEN - 214
7 Response of First-Order RL and RCCircuitsAssessment ProblemsAP 7.1 [a] The circuit for t < 0 is shown below. Note that the inductor behaves like a short circuit, effectively eliminating the 2 resistor from the circuit.First combine the 30 an
Texas A&M - ELEN - 214
8 Natural and Step Responses of RLCCircuitsAssessment ProblemsAP 8.1 [a] 1 1 = , (2RC)2 LC 1 , [b] = 5000 = 2RC s1,2 = -5000 [c] 1 = 20,000, LC therefore C = 500 nF C = 1 Ftherefore 25 106 - therefore(103 )(106 ) = (-5000 j5000) rad/s 20
Texas A&M - ELEN - 214
Sinusoidal Steady State Analysis9Assessment ProblemsAP 9.1 [a] V = 170/-40 V [b] 10 sin(1000t + 20 ) = 10 cos(1000t - 70 ) . I = 10/-70 A[c] I = 5/36.87 + 10/-53.13 = 4 + j3 + 6 - j8 = 10 - j5 = 11.18/-26.57 A [d] sin(20,000t + 30 ) = cos(20,0