K 2 2 K
Author: o | 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 ) / ∑
2025-04-03K − 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 ·
2025-04-05. . , 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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