đối với nghiên cứu này tác giả chọn tiêu chuẩn thang đo phù hợp khi thang đo có hệ số tin cậy cronbach alpha từ 0 6 trở lên
Đề cương chi tiết học phần Kinh tế lượng (Học viện tài chính)
Ngày tải lên :
08/02/2017, 09:15
... 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 quy 4.2 Khoảng tin cậy ... lệch chuẩn hệ số hồi quy ước lượng 1.4 Các tính chất ước lượng bình phương nhỏ Hệ số xác định 2.1 Sai lệch mô hình hồi qui mẫu 2.2 Hệ số xác định Khoảng tin cậy kiểm định giả thuyết hệ số hồi ... 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 phù hợp hàm hồi qui 4.1 Phân tích phương...
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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
... 2 50 75 25 800 3 20 80 32 700 200 70 20 60 0 200 60 20 100 50 1 50 70 200 90 1 50 80 125 60 789 0 20 200 0 100 0 100 0 100 0 100 0 500 1 60 200 30 7 50 200 2 50 45 100 0 300 300 50 7 50 200 200 45 500 1 50 20 ... 200 0 - 500 300 0 -2 50 3 500 -2 50 300 0 +2 50 200 0 +2 50 1 500 27 50 32 50 32 50 22 50 + 500 100 0 1 500 60 0 200 60 20 7 50 2 50 75 25 800 3 20 80 32 700 200 70 20 60 0 200 60 20 100 50 1 50 70 200 90 1 50 80 125 60 ... khấu 10% , có kết NPV A $ 200 0 NPV dự án B $ 500 0 300 − 100 0 r = 300 0 - 100 0 NPVA = 300 − 100 0 0, 1 = 200 0 = 100 0 100 0 − 500 0 = - 500 0 r 0, 1 = 100 00 - 500 0 = 500 0 NPVB = Khi so sánh giá ròng dự án này, ...
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McGraw-HillDictionary of Engineering Second Edition McGraw-Hill 2003 pptx
Ngày tải lên :
27/06/2014, 18:20
... 3 40 and 4 20 F (1 70 and 215ЊC) { odиo klav kyurи ˙ ˙ ¯ ¯ iŋ } autoclave molding [ENG] A method of curing reinforced plastics that uses an autoclave with 50 100 pounds per square inch (345 6 90 ... System .6 30 Special constants 63 4 Electrical and magnetic units 63 5 Dimensional formulas of common quantities .63 5 Internal energy and generalized work 63 6 General ... ¨ 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...
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Boston University/ Met College Administrative Sciences: McGraw-Hill/Irwin doc
Ngày tải lên :
23/03/2014, 03:20
... Accounts Payable 1 2 LF Debit 10 30 14 10 10 21 15 10 13 20 Credit 5 ,00 0 .00 5 ,00 0 .00 7 50. 00 7 50. 00 4 ,00 0 .00 4 ,00 0 .00 7, 200 .00 7, 200 .00 800 .00 800 .00 and any useful supplementary information ... Adjustments Dr Cr 5,4 50 0 5 50 7 50 7, 200 Income Statement Dr Cr 5,4 50 0 5 50 0 7, 200 7 50 2, 200 4 ,00 0 5 ,00 0 0 12, 200 2, 200 4 ,00 0 5 ,00 0 0 12, 200 6 ,00 0 3 ,00 0 4 50 23, 400 Rent expense Depreciation ... 1.ϩ$ 20, 000 ϩ$ 20, 000 Investment 2.Ϫ$ 5 ,00 0 ϩ$7 ,00 0 ϩ$ 2 ,00 0 3.Ϫ$ 1 ,00 0 ϩ$1 ,00 0 4.Ϫ$ 4, 500 Ϫ$ 4, 500 Salaries 5.ϩ$ 5 ,00 0 ϩ$5 ,00 0 ϩ$ 10, 000 Revenues...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 1 pot
Ngày tải lên :
05/08/2014, 20:22
... curve 0 0 .01 32 6. 67 0. 03 13.33 0. 033 16 0. 067 31. 30 0.1 45. 46 0. 133 60 . 26 0. 167 75 . 06 0. 2 89.74 iv 0. 233 0. 267 0. 3 0. 333 0. 367 0. 4 0. 433 0. 467 0. 5 0. 533 0. 567 0. 6 0. 63 3 0. 66 7 0. 7 0. 733 0. 767 0. 8 0. 833 ... A = 0. 002 35 Wb (0. 09 cm ) (0. 05 cm ) = = 0. 522 T 0. 001 56 Wb ( 0. 15 cm )( 0. 05 cm ) = 0. 208 T 0. 000 79 Wb = 0. 1 76 T ( 0. 09 cm )( 0. 05 cm ) A wire is shown in Figure P1 -6 which is carrying 5 .0 A ... )(1 .05 ) ( l3 µr 0 A3 = ) 1.11 m = 90. 1 kA ⋅ t/Wb ( 200 0) 4π × 10 H/m (0. 07 m ) (0. 07 m) ( −7 ) l4 0. 000 5 m = = 77.3 kA ⋅ t/Wb −7 0 A4 4π × 10 H/m (0. 07 m )( 0. 07 m )(1 .05 ) ( l5 µr 0 A5 = ) 0. 37...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 2 ppt
Ngày tải lên :
05/08/2014, 20:22
... 287 W VS ′ 81852 = = 268 W RC 2 50, 000 The efficiency of this transformer is η= 2-3 POUT 16 ,00 0 × 100 % = × 100 % = 96. 6% POUT + PCU + Pcore 16 ,00 0 + 287 + 268 A 100 0-VA 2 30/ 115-V transformer has ... a = 800 0/4 80 = 16. 67 Therefore, the secondary impedances referred to the primary side are RS ′ = a RS = ( 16. 67) ( 0. 05 Ω ) = 13.9 Ω X S ′ = a X S = ( 16. 67) (0. 06 Ω ) = 16. 7 Ω 24 The resulting ... POUT 800 W × 100 % = × 100 % = 93.7% PIN 854 W A 20- kVA 800 0/4 80- V distribution transformer has the following resistances and reactances: RP = 32 Ω X P = 45 Ω RS = 0. 05 Ω X S = 0. 06 Ω RC = 2 50 kΩ...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 3 ppsx
Ngày tải lên :
05/08/2014, 20:22
... I SC 2 60 W = 76. 7° 11 30 V )(1 .00 A ) ( Z EQ = REQ + jX EQ = 11 30 76. 7° = 2 60 + j1 100 Ω The resulting per-unit impedances are REQ = 2 60 Ω = 0. 013 pu 20, 000 Ω X EQ = 1 100 Ω = 0. 055 pu 20, 000 Ω The ... φ ,OC Vφ ,OC θ = − cos−1 = 1. 60 A = 0. 003 33 S 4 80 V POC = − cos −1 VOC I OC 305 W = 66 .6 (4 80 V )(1. 60 A ) YEX = GC − jBM = 0. 003 33∠ − 66 .6 = 0. 001 32 − j 0. 003 06 RC = 1/ GC = 757 Ω X M = 1/ ... POUT + PCU + Pcore = 0. 9 + 0. 0125 + 0. 01 36 = 0. 9 26 and the efficiency of the transformer bank is η= 2-19 POUT 0. 9 × 100 % = × 100 % = 97.2% 0. 9 26 PIN A 20- kVA 20, 000 /4 80- V 60 - Hz distribution transformer...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 4 pot
Ngày tải lên :
05/08/2014, 20:22
... 0. 0 10 + j0 .04 0 + 0. 007 23 + j 0. 0482 + 0. 0 40 + j 0. 1 70 + (1.513 + j1.134 )( − j3. 36) (1.513 + j1.134 ) + ( − j3. 36) Z EQ = 0. 0 10 + j0 .04 0 + 0. 007 88 + j 0. 0525 + 0. 0 40 + j0.1 70 + (2.358 + j0. 109 ) ... T1 1 0 j0 .04 82 0. 0 40 Line + - j0.1 70 T2 1.513 j1.134 (b) L1 With the switch opened, the equivalent impedance of this circuit is Z EQ = 0. 0 10 + j0 .04 0 + 0. 007 23 + j0 .04 82 + 0. 0 40 + j 0. 1 70 + 1.513 ... (1)( 0. 409 ) sin 6 .0 = 0. 0428 SG ,pu = VI = (1) (0. 409 ) = 0. 409 PG = PG ,pu Sbase = ( 0. 407 )( 100 0 kVA ) = 407 kW QG = QG ,pu S base = (0. 0428)( 100 0 kVA ) = 42.8 kVAR SG = SG ,pu S base = (0. 409 ...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 5 ppsx
Ngày tải lên :
05/08/2014, 20:22
... the circuit Since vC (0) = V and vC(∞) = 100 V, it is possible to solve for A and B A = vC(∞) = 100 V A + B = vC (0) = V ⇒ B = - 100 V Therefore, vC (t ) = 100 − 100 e − t 0. 50 V The time at which ... e R2 If we ignore the continuing trickle of current from R1, the time at which iD(t) reaches IH is t2 = − R2C ln I H R2 (0. 000 5 A )(1 500 Ω ) = 5.5 ms = − ( 0. 001 5) ln 30 V VBO Therefore, the period ... degrees Note that % 1/ 60 s = 3 60 degrees for a 60 Hz waveform! theta = 21 60 0 * t(ii); % Calculate the voltage in each phase at each % angle va(ii) = biphase_controller(theta ,0, 90) ; vb(ii) = biphase_controller(theta,-1 20, 90) ;...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 6 pps
Ngày tải lên :
05/08/2014, 20:22
... Poles 10 12 14 4-3 f e = 50 Hz f e = 60 Hz f e = 400 Hz 300 0 r/min 1 500 r/min 100 0 r/min 7 50 r/min 60 0 r/min 500 r/min 428 .6 r/min 3 60 0 r/min 1 800 r/min 1 200 r/min 900 r/min 7 20 r/min 60 0 r/min ... r/min 514.3 r/min 2 400 0 r/min 1 200 0 r/min 800 0 r/min 60 0 0 r/min 4 800 r/min 400 0 r/min 3429 r/min A three-phase four-pole winding is installed in 12 slots on a stator There are 40 turns of wire in ... plot(real(s_curve),imag(s_curve),'b','LineWidth',2 .0) ; hold on; plot(real(r_curve),imag(r_curve),'r ','LineWidth',2 .0) ; % Add x and y axes 111 plot( [-1 500 1 500 ], [0 0],'k'); plot( [0, 0],[-1 500 1 500 ],'k'); % Set titles and...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 7 doc
Ngày tải lên :
05/08/2014, 20:22
... ( 100 kW = ( 0. 1 MW/Hz ) 60 . 5 Hz − f sys + (0. 15 MW/Hz ) 60 . 0 Hz − fsys ) 100 kW = 60 5 0 kW − ( 0. 10 MW/Hz ) fsys + 900 0 kW − (0. 15 MW/Hz ) fsys (0. 25 MW/Hz ) f sys = 60 5 0 kW + 900 0 kW − 100 kW ... = nnl − nfl 36 30 r/min − 35 70 r/min × 100 % = × 100 % = 1 .68 % nfl 35 70 r/min The speed droop of generator is given by SD2 = (b) nnl − nfl 1 800 r/min − 1785 r/min × 100 % = × 100 % = 0. 84% nfl 1785 ... = (1.5) 61 .0 − f sys + (1 .67 6) 61 .5 − f sys + (1. 961 ) 60 . 5 − f sys ) MW = 91.5 − 1.5f sys + 103 .07 − 1 .67 6f sys + 118 .64 − 1. 961 f sys 5.137 fsys = 3 06 .2 f sys = 59 .61 Hz The power supplied by...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 8 ppt
Ngày tải lên :
05/08/2014, 20:22
... below Synchronous Motor V-Curve 2 60 2 50 A I (A) 2 40 2 30 2 20 2 10 200 3 50 400 4 50 500 5 50 60 0 6 50 700 E (V) A 6- 3 A 2 300 -V 100 0-hp 0. 8-PF leading 60 - Hz two-pole Y-connected synchronous motor has ... = (0. 85)( 100 MVA ) = 85 MW and 1 20 f e 1 20 ( 50 Hz ) = = 300 0 r/min P nsync = the applied torque would be τ app = τ ind = 5-23 85 ,00 0 ,00 0 W = 2 70, 00 0 N ⋅ m ( 300 0 r/min )(2π rad/r )(1 min/ 60 s) ... 4 80 V )( 400 A )( 0. 9 ) = 299 kW Q1 = VT I L sin θ = ( 4 80 V )( 400 A ) sin cos −1 ( 0. 9 ) = 145 kVAR P2 = VT I L cos θ = ( 4 80 V )( 200 A )( 0. 72 ) = 1 20 kW Q2 = VT I L sin θ = ( 4 80 V )( 200 ...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 9 pot
Ngày tải lên :
05/08/2014, 20:22
... nsync × 100 % = 1 800 − 17 20 × 100 % = 4.44% 1 800 f r ,fl = sf e = ( 0. 0444 )( 60 Hz ) = 2 .67 Hz The speed regulation is SR = 7-7 nnl − nfl 17 90 − 17 20 × 100 % = × 100 % = 4.1% 17 20 nfl A 208 -V, two-pole, ... s = 0. 25 (0. 066 7) = 0. 0 167 The resulting speed is nm = (1 − s ) nsync = (1 − 0. 0 167 )( 900 r/min ) = 885 r/min (d) The electrical frequency at ¼ load is f r = sf e = ( 0. 0 167 )( 60 Hz ) = 1 .00 Hz ... machine is 1 800 r/min The slip and electrical frequency at noload conditions is nsync − nnl snl = nsync × 100 % = 1 800 − 17 90 × 100 % = 0. 56% 1 800 f r ,nl = sf e = ( 0. 005 6) ( 60 Hz ) = 0. 33 Hz The...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 10 pps
Ngày tải lên :
05/08/2014, 20:22
... is shown below: 177 Induction Motor Torque-Speed Characteristic 60 50 τ ind 40 30 20 10 0 500 100 0 1 500 200 0 n 2 500 300 0 3 500 400 0 m (b) A MATLAB program to calculate the output-power versus speed ... 500 300 200 100 2 700 27 50 2 800 n m 1 90 28 50 (r/min) 2 900 29 50 300 0 This machine is rated at 75 kW It produces an output power of 75 kW at 3.1% slip, or a speed of 2 907 r/min 7-18 A 208 -V, 60 ... jX ) ( j8 .64 Ω )( 0. 075 Ω + j 0. 204 Ω ) = = 0. 0731 + j 0. 1994 Ω = 0. 212 69 .9° Ω R1 + j ( X + X M ) 0. 075 Ω + j ( 0. 204 Ω + 8 .64 Ω) VTH = jX M ( j8 .64 Ω ) Vφ = (254 0 V ) = 248 0. 05° V R1 +...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 1 pptx
Ngày tải lên :
05/08/2014, 20:22
... Characteristic 4 50 R2 = 0. 005 9 ohms R2 = 0. 172 ohms 400 3 50 300 τ ind 2 50 200 1 50 100 50 1 60 0 16 20 16 40 16 60 16 80 1 700 n 17 20 17 40 17 60 17 80 1 800 m (b) The slip at pullout torque can be found by calculating ... of 0. 04 Ω Therefore, the total resistance R A is RA = (0. 04 Ω + 0. 04 Ω + 0. 04 Ω + 0. 04 Ω ) (0. 04 Ω + 0. 04 Ω + 0. 04 Ω + 0. 04 Ω ) 0. 04 Ω + 0. 04 Ω + 0. 04 Ω + 0. 04 Ω + 0. 04 Ω + 0. 04 Ω + 0. 04 Ω + 0. 04 ... M at starting conditions (s = 1 .0) is: 1 ZF = = = 0. 167 + j0.415 = 0. 448 68 .0 Ω 1 1 + + jX M Z j 30 Ω 0. 172 + j 0. 42 The phase voltage is 4 60 / = 266 V, so line current I L is Vφ 266 0 V = IL...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 2 ppt
Ngày tải lên :
05/08/2014, 20:22
... 1 200 No armature reaction With armature reaction 11 50 nm (r/min) 1 100 105 0 100 0 9 50 900 8 50 800 20 40 60 τ ind 80 100 1 20 (N-m) For Problems 9-8 and 9-9, the shunt dc motor is reconnected separately ... (2 40 W ) = 105 0 r/min (2 40 W ) = 483 W Therefore, the output power is POUT = Pconv − Pmech − Pcore = 3759 W − 483 W − 200 W = 307 6 W and the efficiency is η= (c) 307 6 W POUT × 100 % = × 100 % = ... Ao of 2 90 V at a speed no of 1 200 r/min Therefore, n= EA E Ao no = 2 16. 2 V (1 200 r/min ) = 895 r/min 2 90 V The speed regulation is SR = 1 06 3 r/min − 895 r/min nnl − nfl × 100 % = × 100 % = 18.8%...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 3 docx
Ngày tải lên :
05/08/2014, 20:22
... vs field current'); axis ( [0 1 50] ); set(gca,'YTick', [0 10 20 30 40 50 60 70 80 90 100 1 10 1 20 1 30 1 40 1 50] ') set(gca,'XTick', [0 0.5 1 .0 1.5 2 .0 2.5 3 .0 3.5 4 .0 4.5 5 .0] ') legend ('Ea line','Vt ... FAR = ( 100 0 turns )(5 A ) − 400 A ⋅ turns = 4 60 0 A ⋅ turns or I F * = Fnet / N F = 4 60 0 A ⋅ turns / 100 0 turns = 4 .6 A The equivalent internal generated voltage E Ao of the generator at 1 800 r/min ... n= (c) EA 2 26. 1 V no = (1 200 r/min ) = 1 06 0 r/min E Ao 2 56 V If Radj is maximum at no-load conditions, the total resistance is 500 Ω, and IF = VT 2 40 V = = 0. 48 A RF + Radj 200 Ω + 300 Ω This field...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 4 pptx
Ngày tải lên :
05/08/2014, 20:22
... vs field current'); axis ( [0 1 50] ); set(gca,'YTick', [0 10 20 30 40 50 60 70 80 90 100 1 10 1 20 1 30 1 40 1 50] ') set(gca,'XTick', [0 0.5 1 .0 1.5 2 .0 2.5 3 .0 3.5 4 .0 4.5 5 .0] ') legend ('Ea line','Vt ... jX + jX M ( 100 + j 2. 40) ( j 60 ) = 2. 96 + 100 + j 2. 40 + j 60 R2 / ( s ) jX j 2. 46 Ω ( jX M ) R2 / ( − s ) + jX + jX M (1.282 + j 2. 40 )( j 60 ) = 1. 90 + j 2.37 Ω 1.282 + j 2. 40 + j 60 The input ... jX M ( 100 + j 2. 40 )( j 60 ) = 28.91 + 100 + j 2. 40 + j 60 j 43.83 Ω 271 60 s = 1 .68 N ⋅ m R2 / ( s ) ZB = ZB = (a) jX ( jX M ) R2 / ( − s ) + jX + jX M (1.282 + j 2. 40) ( j 60 ) = 1.1 70 + j 2.331...
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