kết quả của hệ số tin cậy cronbach alpha đối với thang đo các yếu tố ảnh hưởng đến quyết định gửi tiền tiết kiệm của khách hàng cá nhân tại các nhtm là

Đề cương chi tiết học phần Kinh tế lượng (Học viện tài chính)

Đề cương chi tiết học phần Kinh tế lượng (Học viện tài chính)

Ngày tải lên : 08/02/2017, 09:15
... 2.2 Hệ số xác định Khoảng tin cậy kiểm định giả thuyết hệ số hồi qui 3.1 Khoảng tin cậy số hồi quy 3.2 Kiểm định giả thuyết hệ số hồi quy 3.3 Khoảng tin cậy σ2 3.4 Kiểm định giả thuyết σ2 Kiểm định ... phương nhỏ Hệ số xác định bội 3.1 Hệ số xác định bội R2 3.2 Hệ số xác định bội hiệu chỉnh R Khoảng tin cậy kiểm định giả thuyết mô hình hồi qui bội 4.1 Khoảng tin cậy kiểm định giả thuyết hệ số hồi ... quan hệ biến định lượng mở rộng biến định tính - Ước lượng các hệ số đại lượng mô hình hồi quy phương pháp bình phương nhỏ Từ giá trị ước lượng mẫu số liệu, thực suy diễn thống kê mối quan hệ kinh...
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Chương trình Giảng dạy Kinh tế Fulbright Niên khoá 2004  2005: Thẩm định đầu tư phát triển

Chương trình Giảng dạy Kinh tế Fulbright Niên khoá 2004 2005: Thẩm định đầu tư phát triển

Ngày tải lên : 27/07/2014, 11:02
... theo cách xử lý chi phí hành Ngược lại, giá ròng dự án khơng nhạy cảm với cách xử lý chi phí kế tốn viên Tiếc thay, thẩm định dự án có nhiều định tùy tiện cách cân đối lợi ích với chi phí định ảnh ... Thẩm định đầu tư phát triển Bài đọc Sách hướng dẫn Ch Chiến lược thẩm định dự án đầu tư cơng cộng Các đối tượng hưởng lợi nhận lợi ích theo cách đối tượng chịu chi phí dự án phải trả theo cách ... (c) Thẩm định đầu tư phát triển Bài đọc Sách hướng dẫn Ch Chiết khấu tiêu chuẩn để đánh giá đầu tư Các yếu tố ảnh hưởng đến suất chiết khấu dự án khu vực cơng Đối với đầu tư khu vực tư nhân, suất...
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McGraw-HillDictionary of Engineering Second Edition McGraw-Hill 2003 pptx

McGraw-HillDictionary of Engineering Second Edition McGraw-Hill 2003 pptx

Ngày tải lên : 27/06/2014, 18:20
... ¨ actinogram [ENG] The record of heat from a source, such as the sun, as detected by a recording actinometer { ak tin ə gram } actinograph [ENG] A recording actinometer { ak tin ə graf } actinometer ... accent lighting [CIV ENG] Directional lighting which highlights an object or attracts attention to a particular area { akиsent lıdиiŋ } ¯ acceptability [ENG] State or condition of meeting minimum ... effectiveness of a piece of equipment without changing its basic function; may be used for testing, adjusting, calibrating, recording, or other purposes { ak sesиəиre } ¯ access road [CIV ENG] A route,...
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Boston University/ Met College Administrative Sciences: McGraw-Hill/Irwin  doc

Boston University/ Met College Administrative Sciences: McGraw-Hill/Irwin  doc

Ngày tải lên : 23/03/2014, 03:20
... information Consists of Accounting information Nonaccounting information Consists of Operating information Financial accounting Management accounting Tax accounting tained for each employee showing ... 0−390−30185−X Accounting Contents Anthony−Hawkins−Merchant • Accounting: Text and Cases, Tenth Edition I Financial Accounting 1 The Nature and Purpose of Accounting Accounting Records and Systems ... provides much of the basic data for management accounting, financial accounting, and tax accounting Financial Accounting Information Financial accounting information is intended both for managers and...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 1 pot

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 1 pot

Ngày tải lên : 05/08/2014, 20:22
... each problem in the book This structure should make it easier to copy pages from the manual for posting after problems have been assigned Many of the problems in Chapters 2, 5, 6, and require that ... students before assigning homework problems (Note that this error will be corrected at the second printing, so it may not be present in your student’s books.) v The solutions in this manual have been ... the book’s Web site, which is http://www.mhhe.com/engcs/electrical/chapman/ I am also contemplating a homework problem refresh, with additional problems added on the book’s Web site midway through...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 2 ppt

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 2 ppt

Ngày tải lên : 05/08/2014, 20:22
... circumstances? 19 SOLUTION (a) The current in the bar at starting is V 100 V i= B = = 400 A R 0.25 Ω Therefore, the force on the bar at starting is F = i ( l × B) = ( 400 A )(1 m )(0.5 T ) = 200 N, ... length l = 1.0 m, and a battery voltage of 100 V (a) What is the initial force on the bar at starting? What is the initial current flow? (b) What is the no-load steady-state speed of the bar? (c) ... α cos β + sin α sin β The result is p (t ) = VI [ cosθ + cos2ω t cos θ + sin 2ω t sin θ ] Collecting terms yields the final result: p (t ) = VI cosθ (1 + cos 2ω t ) + VI sin θ sin 2ω t 1-21 A...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 3 ppsx

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 3 ppsx

Ngày tải lên : 05/08/2014, 20:22
... apparent power rating of each transformer is 1/3 of the total apparent power rating of the three-phase transformer For the open-∆ and open-Y—open-∆ connections, the apparent power rating is a bit ... 86.6% of the total apparent 38 power rating of the two transformers, implying that the apparent power rating of each transformer must be 231 kVA The ratings for each transformer in the bank for ... off; The resulting plot of secondary voltage versus load is shown below: 2-9 A three-phase transformer bank is to handle 600 kVA and have a 34.5/13.8-kV voltage ratio Find the rating of each individual...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 4 pot

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 4 pot

Ngày tải lên : 05/08/2014, 20:22
... connections when it is used as a 600/480-V step-down autotransformer (c) What is the kilovoltampere rating of this transformer when it is used in the autotransformer connection? (d) Answer the questions ... SC 26 W (10.0 V )(10.6 A ) = 75.8° Z EQ = REQ + jX EQ = 0.943∠75.8° = 0.231 + j 0.915 Ω The resulting per-unit impedances are REQ = 0.231 Ω = 0.010 pu 23.04 Ω X EQ = 0.915 Ω = 0.0397 pu 23.04 Ω ... operation is 58 + NSE + 600 V NC (c) VC + 480 V - - When used as an autotransformer, the kVA rating of this transformer becomes: SIO = (d) + VSE N C + N SE +1 SW = (10 kVA) = 50 kVA N SE As an...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 5 ppsx

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 5 ppsx

Ngày tải lên : 05/08/2014, 20:22
... capacitors C1 and C2 will force SCR2 and SCR3 to turn OFF The cycle continues in this fashion Capacitors C1 and C2 are called commutating capacitors Their purpose is to force one set of SCRs to turn ... / 12 9T / 12 11T / 12 T / 12 c a a b b c c b b c c a a b 88 Conducting SCR (Positive) SCR3 SCR1 SCR1 SCR2 SCR2 SCR3 SCR3 Conducting SCR (Negative) SCR5 SCR5 SCR6 SCR6 SCR4 SCR4 SCR5 Triggered ... SCR2 is triggered When SCR2 turns ON, capacitor C applies a reverse-biased voltage to SCR1, shutting it off Current then flow through the capacitor, SCR2, and the load as shown below This current...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 6 pps

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 6 pps

Ngày tải lên : 05/08/2014, 20:22
... operating at frequencies of 50, 60, and 400 Hz SOLUTION The equation relating the speed of magnetic field rotation to the number of poles and electrical frequency is nm = 120 f e P The resulting ... power system for lighting? SOLUTION (a) The equation relating the speed of magnetic field rotation to the number of poles and electrical frequency is nm = 120 f e P The resulting table is Number ... Bcc; % Calculate a circle representing the expected maximum % value of Bnet circle = 1.5 * (cos(w*t) + j*sin(w*t)); % Plot the magnitude and direction of the resulting magnetic % fields Note that...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 7 doc

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 7 doc

Ngày tải lên : 05/08/2014, 20:22
... with the existing power system? What is the generator’s rate of shaft rotation after paralleling occurs? 118 (b) If the generator is connected to the power system and is initially floating on the ... power system and is initially floating on the line, sketch the resulting magnetic fields and phasor diagram (c) The governor setting on the diesel is now increased Show both by means of house diagrams ... connecting the two systems together should be shut when the above conditions are met and the generator is in phase with the power system After paralleling, the generator’s shaft will be rotating...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 8 ppt

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 8 ppt

Ngày tải lên : 05/08/2014, 20:22
... generator is directly proportional to the speed of the generator, the voltage rating (and hence the apparent power rating) of the generator will be reduced by a factor of 5/6 VT ,rated = S rated ... decreased 5-25 A generating station for a power system consists of four 120-MVA 15-kV 0.85-PF-lagging synchronous generators with identical speed droop characteristics operating in parallel The ... + RAI A + jX S I A E A = 7621∠0° + j ( 0.9 Ω)(5249 ∠ − 36.87° A ) E A = 11,120∠19.9° V The resulting voltage regulation is VR = 11,120 − 7621 × 100% = 45.9% 7621 (b) If the generator is to be...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 9 pot

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 9 pot

Ngày tải lên : 05/08/2014, 20:22
... supplied or consumed by the machine under the conditions in part (a) Is the machine operating within its ratings under these circumstances? (c) If E A = 470∠-12° V and Vφ = 440∠0° V, is this machine ... supplied or consumed by the machine under the conditions in part (c) Is the machine operating within its ratings under these circumstances? SOLUTION (a) This machine is a generator supplying real ... × 100% = 6.67% 900 The motor is operating in the linear region of its torque-speed curve, so the slip at ¼ load will be s = 0.25(0.0667) = 0.0167 The resulting speed is nm = (1 − s ) nsync = (1...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 10 pps

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 10 pps

Ngày tải lên : 05/08/2014, 20:22
... at starting conditions (when the shaft is not moving)? Plot the torque-speed characteristic of this motor with the additional resistance inserted SOLUTION To get the maximum torque at starting, ... (0.4016 Ω + 0.410 Ω ) R2 = 0.833 Ω Since the existing resistance is 0.120 Ω, an additional 0.713 Ω must be added to the rotor circuit The resulting torque-speed characteristic is: 179 7-11 If ... resulting equivalent circuit is shown below IA R1 + jX1 0.075 Ω j0.204 Ω j8.64 Ω Vφ jX2 R2 j0.204 Ω 0.065 Ω jXM 1− s  R2    s  1.56 Ω - The slip at pullout torque is found by calculating...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 1 pptx

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 1 pptx

Ngày tải lên : 05/08/2014, 20:22
... starting, what is the maximum starting current that the controller must be designed to handle? (c) If a 1.25:1 step-down autotransformer starter is used during starting, what is the maximum starting ... (or armature current) at starting is to get the equivalent impedance Z F of the rotor circuit in parallel with jX M at starting conditions, and then calculate the starting current as the phase voltage ... motor starting, which is less than the 30% sag with case of across-the-line starting 7-21 In this chapter, we learned that a step-down autotransformer could be used to reduce the starting current...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 2 ppt

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 2 ppt

Ngày tải lên : 05/08/2014, 20:22
... r/min ) = 1447 r/min 199 V What is the starting current of this machine if it is started by connecting it directly to the power supply VT ? How does this starting current compare to the full-load ... Calculate the resulting internal generated voltage at % 1200 r/min by interpolating the motor's magnetization % curve e_a0 = interp1(if_values,ea_values,i_f); % Calculate the resulting speed from ... Calculate the resulting internal generated voltage at % 1200 r/min by interpolating the motor's magnetization % curve e_a0 = interp1(if_values,ea_values,i_f); % Calculate the resulting speed from...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 3 docx

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 3 docx

Ngày tải lên : 05/08/2014, 20:22
... Calculate the resulting internal generated voltage at % 1800 r/min by interpolating the motor's magnetization % curve e_a0 = interp1(if_values,ea_values,i_f); 237 % Calculate the resulting speed from ... Calculate the resulting internal generated voltage at % 1800 r/min by interpolating the motor's magnetization % curve e_a0 = interp1(if_values,ea_values,i_f); % Calculate the resulting speed from ... Calculate the resulting internal generated voltage at % 1800 r/min by interpolating the motor's magnetization % curve e_a0 = interp1(if_values,ea_values,i_f); % Calculate the resulting speed from...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 4 pptx

(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 4 pptx

Ngày tải lên : 05/08/2014, 20:22
... accelerating through 400 r/min How much induced torque will the motor be able to produce on its main 272 winding alone? Assuming that the rotational losses are still 51 W, will this motor continue ... might just be able to keep on accelerating slowly—it is a close thing either way 273 SOLUTION This problem is best solved with MATLAB, since it involves calculating the torque-speed values at many ... following questions about this machine, assuming no armature reaction (a) If the generator is operating at no load, what is its terminal voltage? (b) If the generator has an armature current of 20...
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