2 K
Author: b | 2025-04-24
= 0 k 2 k 2 k 3 k k 2 2 k 2 7 k 2 k = 1 ⇒ 10 k 2 9 k = 1 ⇒ 10 k 2 9 k − 1 = 0 ⇒ (10 k − 1) (k 1) = 0 ⇒ k = − 1, 1 10 It is known that probability of any observation must always
Factor k^2-k-2
T p 1 ) T p 2 / ( 1 − T p 2 ) give 95% confidence limits for the odds ratio in the absence of covariates. 3.2. The Common Odds RatioConsider K independent studies (or strata from the same study), where from the kth study, we have observations for two independent binomial random variables X 1 k and X 2 k with respective success probabilities p 1 k and p 2 k , and respective sample sizes n 1 k and n 2 k , k = 1, 2, ...., K. Thus, the odds ratio from the kth study is δ k = p 1 k / ( 1 − p 1 k ) p 2 k / ( 1 − p 2 k ) , k = 1, 2, ...., K. Assuming that the odds ratio is the same across the K studies, we have δ 1 = δ 2 = . . . . = δ K = δ (say). 3.2.1. An Approximate GPQ for the Common Odds RatioAn approximate GPQ for each δ k , to be denoted by T δ k , can be constructed from the kth study, proceeding as mentioned in Section 3.1. We now combine these GPQs in order to obtain an approximate GPQ for the common odds ratio δ. For this, we propose a weighted average of the study-specific GPQs on the log scale. The weights that we shall use are motivated as follows. For i = 1, 2, if p ^ i k denote sample proportions from the kth study, and if δ ^ k = p ^ 1 k / ( 1 − p ^ 1 k ) p ^ 2 k / ( 1 − p ^ 2 k ) , k = 1, 2, ...., K, then using the delta method, an approximate variance of log ( δ ^ k ) , say 1 / w k , is given by: 1 / w k = 1 n 1 k p 1 k + 0.5 + 1 n 1 k ( 1 − p 1 k ) + 0.5 + 1 n 2 k p 2 k + 0.5 + 1 n 2 k ( 1 − p 2 k ) + 0.5 , where we have also used a continuity correction. Noting that log ( δ ) = ∑ k = 1 K w k log ( δ k ) / ∑ k = 1 K w k , an approximate GPQ T δ for the common odds ratio can be obtained from log ( T δ ) = ∑ k = 1 K T w k log ( T δ k ) / ∑. = 0 k 2 k 2 k 3 k k 2 2 k 2 7 k 2 k = 1 ⇒ 10 k 2 9 k = 1 ⇒ 10 k 2 9 k − 1 = 0 ⇒ (10 k − 1) (k 1) = 0 ⇒ k = − 1, 1 10 It is known that probability of any observation must always k 2 (k 1) 2 (k 1) 3 = (k 1) 2 (k 2) 2. Multiply all terms by 4: k 2 (k 1) 2 4(k 1) 3 = (k 1) 2 (k 2) 2. All terms have a common factor (k 1) 2, so it can be canceled: k 2 4(k 1) = (k 2) 2. And simplify: k 2 4k 4 = k 2 4k 4. They are the same! So it is true. So: 1 3 2 3 3 3 (k 1) 3 = (k 1 If K 2 4 K 8, 2 K 2 3 K 6 and 3 K 2 4 K 4 are any 3 consecutive terms of A. P. View Solution. Q4. Question 11 Determine k, so that k 2 4 k 8, 2 k 2 3 k 6 a n d 3 k 2 4 k We know that 1 3 2 3 3 3 k 3 = k 2 (k 1) 2 (the assumption above), so we can do a replacement for all but the last term: k 2 (k 1) 2 (k 1) 3 = (k 1) 2 (k 2) 2. Multiply all terms by 4: k 2 (k 1) 2 4(k 1) 3 = (k 1) 2 A chemical reaction was carried out at 300K and 280K. The rate conatants were found to be K 1 and K 2 respectively. Then. 1)K 2 = 4 K 1. 2)K 2 =2 K 1. 3)K 2 = 0.5 K 1. 4)K 2 = 0.25 K 1 K − 1 ) + a 2 · φ CO ( k − 1 ) + a 3 · φ CO 2 ( k − 1 ) + a 4 · φ CH 4 ( k − 1 ) + a 5 · T max ( k − 1 ) , (8) u 2 ≡ V O 2 ( k ) = a 0 + a 1 · V air ( k − 1 ) V O 2 ( k − 1 ) + a 2 · φ CO ( k − 1 ) + a 3 · φ CO 2 ( k − 1 ) + a 4 · φ CH 4 ( k − 1 ) + a 5 · T max ( k − 1 ) . (9) The third model type aims to achieve the necessary temperature for syngas production. The models regarding the ratio of gasification agents and the maximum temperature T m a x in the channel are in the linear and quadratic form. The structure of the model is as the following [25]: u 1 ≡ V air ( k ) = a 0 + a 1 · V air ( k − 1 ) V O 2 ( k − 1 ) + a 2 · T max ( k − 1 ) + a 3 T max 2 ( k − 1 ) , (10) u 2 ≡ V O 2 ( k ) = a 0 + a 1 ·Comments
T p 1 ) T p 2 / ( 1 − T p 2 ) give 95% confidence limits for the odds ratio in the absence of covariates. 3.2. The Common Odds RatioConsider K independent studies (or strata from the same study), where from the kth study, we have observations for two independent binomial random variables X 1 k and X 2 k with respective success probabilities p 1 k and p 2 k , and respective sample sizes n 1 k and n 2 k , k = 1, 2, ...., K. Thus, the odds ratio from the kth study is δ k = p 1 k / ( 1 − p 1 k ) p 2 k / ( 1 − p 2 k ) , k = 1, 2, ...., K. Assuming that the odds ratio is the same across the K studies, we have δ 1 = δ 2 = . . . . = δ K = δ (say). 3.2.1. An Approximate GPQ for the Common Odds RatioAn approximate GPQ for each δ k , to be denoted by T δ k , can be constructed from the kth study, proceeding as mentioned in Section 3.1. We now combine these GPQs in order to obtain an approximate GPQ for the common odds ratio δ. For this, we propose a weighted average of the study-specific GPQs on the log scale. The weights that we shall use are motivated as follows. For i = 1, 2, if p ^ i k denote sample proportions from the kth study, and if δ ^ k = p ^ 1 k / ( 1 − p ^ 1 k ) p ^ 2 k / ( 1 − p ^ 2 k ) , k = 1, 2, ...., K, then using the delta method, an approximate variance of log ( δ ^ k ) , say 1 / w k , is given by: 1 / w k = 1 n 1 k p 1 k + 0.5 + 1 n 1 k ( 1 − p 1 k ) + 0.5 + 1 n 2 k p 2 k + 0.5 + 1 n 2 k ( 1 − p 2 k ) + 0.5 , where we have also used a continuity correction. Noting that log ( δ ) = ∑ k = 1 K w k log ( δ k ) / ∑ k = 1 K w k , an approximate GPQ T δ for the common odds ratio can be obtained from log ( T δ ) = ∑ k = 1 K T w k log ( T δ k ) / ∑
2025-03-28K − 1 ) + a 2 · φ CO ( k − 1 ) + a 3 · φ CO 2 ( k − 1 ) + a 4 · φ CH 4 ( k − 1 ) + a 5 · T max ( k − 1 ) , (8) u 2 ≡ V O 2 ( k ) = a 0 + a 1 · V air ( k − 1 ) V O 2 ( k − 1 ) + a 2 · φ CO ( k − 1 ) + a 3 · φ CO 2 ( k − 1 ) + a 4 · φ CH 4 ( k − 1 ) + a 5 · T max ( k − 1 ) . (9) The third model type aims to achieve the necessary temperature for syngas production. The models regarding the ratio of gasification agents and the maximum temperature T m a x in the channel are in the linear and quadratic form. The structure of the model is as the following [25]: u 1 ≡ V air ( k ) = a 0 + a 1 · V air ( k − 1 ) V O 2 ( k − 1 ) + a 2 · T max ( k − 1 ) + a 3 T max 2 ( k − 1 ) , (10) u 2 ≡ V O 2 ( k ) = a 0 + a 1 ·
2025-04-12. . , K [ 31 ] (reduced result).1: A ← ( K [ 31 ] & 0 b 1 ) ≪ 6 2: B ← ( K [ 31 ] & 0 b 10 ) ≪ 5 3: C ← ( K [ 31 ] & 0 b 1111111 ) 4: K [ 16 ] ← K [ 16 ] ⨁ ( ( A ⨁ B ⨁ C ) ≪ 1 ) 5: K [ 8 ] ← K [ 8 ] ⨁ K [ 24 ] ⨁ ( ( K [ 23 ] ≪ 7 ) ∣ ( K [ 24 ] ≫ 1 ) ) ⨁ ( ( K [ 23 ] ≪ 6 ) ∣ ( ( K [ 24 ] ≫ 2 ) ) ⨁ ( ( K [ 23 ] ≪ 1 ) ∣ ( K [ 24 ] ≫ 7 ) ) 6: K [ 0 ] ← K [ 0 ] ⨁ K [ 16 ] ⨁ ( K [ 16 ] ≫ 2 ) ⨁ ( K [ 16 ] ≫ 7 ) 7:forl = 1 to 7 do8: K [ i + 8 ] ← K [ i + 8 ] ⨁ K [ i + 24 ] ⨁ ( ( K [ i + 23 ] ≪ 7 ) ∣ ( K [ i + 24 ] ≫ 1 ) ) ⨁ ( ( K [ i + 23 ] ≪ 6 ) ∣ ( K [ i + 24 ] ≫ 2 ) ) ⨁ ( ( K [ i + 23 ] ≪ 1 ) ∣ ( K [ i + 24 ] ≫ 7 ) ) 9: K [ i ] ← K [ i ] ⨁ K [ i + 16 ] ⨁ ( ( K [ i + 15 ] ≪ 7 ) ∣ ( K [ i + 16 ] ≫ 1 ) ) ⨁ ( ( K [ i + 15 ] ≪ 6 ) ∣ ( K [ i + 16 ] ≫ 2 ) ) ⨁ ( ( K [
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2025-04-12+ K i K f ) ] G 3 = L m m ema + m el L m + J m L m η ema k ema 2 R G 2 = η ema k ema 2 R ( J m R m + J m K i K f + B fm L m ) + B fr L m + ( m ema + m el ) ( R m + K i K f ) G 1 = η ema k ema 2 R ( C T C E + B fm R m + K i K f B fm ) + B fr ( R m + K i K f ) + S t L m G 0 = S t ( R m + K i K f ) By dividing both the numerator and the denominator on the right side of Equation (11) by the coefficient of u, we obtain: F ema = u − ( F 3 s 3 + F 2 s 2 + F 1 s ) x a ( s 2 m el + S t ) R k ema η ema K i K v C T [ ( s 2 m el + S t ) R k ema η ema K i K v C T ] ( G 3 s 3 + G 2 s 2 + G 1 s + G 0 ) (12) It can be seen from Figure 13 that to suppress the disturbance force, the feedforward correction element should be used to counteract the part following u in the numerator of Equation (12): G T = F 3 s 3 + F 2 s 2 + F 1 s ( s 2 m el + S t ) R k ema η ema K i K v C T (13) The simulation model of the DEMA force servo system introduced with a PID controller and a feedforward correction element is shown in Figure 14 [48,49,50].As shown in Figure 14, we added several modules in the model to simulate the sampling and data processing of the processor, making the simulation model more realistic. In order to make the simulation model easier to understand and operate, we use PMSM and planetary-roller screw pairs as super components, and integrate feedforward and its sampling and data processing into the super components.The disturbance of
2025-04-09− 1 ) + a 2 · φ CO ( k − 1 ) + a 3 · φ CO 2 ( k − 1 ) + a 4 · φ CH 4 ( k − 1 ) + a 5 · T ( k − 1 ) , (6) u 2 ≡ V o 2 ( k ) = a 0 + a 1 · V O 2 ( k − 1 ) + a 2 · φ CO ( k − 1 ) + a 3 · φ CO 2 ( k − 1 ) + a 4 · φ CH 4 ( k − 1 ) + a 5 · T ( k − 1 ) , (7) where k represents the control step of the sampling period τ 0 ; V a i r and V O 2 represent the flow rates of injected air and oxygen to the mixture (m 3 /h); φ i is the concentration of CO, CO 2 , and CH 4 in the syngas (%); and T represents the coal temperature in the gasification channel ( ∘ C).The second type of model refers to the ratio of the flow rates of gasification agents and the highest temperature in the gasification channel T m a x . The structure of this model is the following [25]: u 1 ≡ V air ( k ) = a 0 + a 1 · V air ( k − 1 ) V O 2 (
2025-03-25