Спутник технологический. H2 sat. [Редактировать]

Н2 (другое название Ho‘oponopono-2), массой 3,5 кг, создан специалистами Гавайского университета и предназначен для калибровки наземных радаров.

Дополнительная классификация

#Наименования
1Страна оператор(владелец) - США
2Страна производитель - США

Технические характеристики

#ХарактеристикаЗначение
1Масса, кг3.5

Информация об удачном запуске

#ХарактеристикаЗначение
1Космодром Уоллопс
2Дата пуска2013-11-20
3Полезная нагрузка 1xKySat-2
4Полезная нагрузка 1xSTPSat-3
5Полезная нагрузка 1xPrometheus 1A
6Полезная нагрузка 1xPrometheus 1B
7Полезная нагрузка 1xPrometheus 2A
8Полезная нагрузка 1xPrometheus 2B
9Полезная нагрузка 1xPrometheus 3A
10Полезная нагрузка 1xPrometheus 3B
11Полезная нагрузка 1xPrometheus 4A
12Полезная нагрузка 1xPrometheus 4B
13Полезная нагрузка 1xHorus
14Полезная нагрузка 1xORSES
15Полезная нагрузка 1xORS Tech 1
16Полезная нагрузка 1xORS Tech 2
17Полезная нагрузка 1xSENSE 1
18Полезная нагрузка 1xSENSE 2
19Полезная нагрузка 1xH2 sat
20Полезная нагрузка 1xDragonSat-1
21Полезная нагрузка 1xNPS-SCAT
22Полезная нагрузка 1xFirefly
23Полезная нагрузка 1xSPA-1
24Полезная нагрузка 1xChargerSat
25Полезная нагрузка 1xLunar
26Полезная нагрузка 1xCOPPER
27Полезная нагрузка 1xBlack Knight-1
28Полезная нагрузка 1xSwampSat
29Полезная нагрузка 1xCAPE-2
30Полезная нагрузка 1xTJ3Sat
31Полезная нагрузка 1xPhoneSat 2.0
32Ракета-носитель 1xМинотавр 1

Найдено 1000 документов по запросу «H2 sat». [Перейти к поиску]


Дата загрузки: 2017-06-15
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... Weight Relationships 1 - 15 Common Relationships W  W ( 1  w) s V  V ( 1  e) s d  m 1 w    sat   w d  G w s 1 e e n n 1n W – Total Weight V – Total Volume w – water... S – Saturation wsat – Saturated Moisture Content e 1 e w G s e S S w w sat w sat  w d  1 G s  sat   d ( 1  wsat ) 1 - 16 Specific Gravity 1 - 17... u=h2•w Where w is the unit weight of water (62.4 pcf) ’  - u = (h1•m + h2•sat) - h2•w = h1•m +h2(sat - w) = h1•m + h2•’ sat  ’  sat -  w  Gs  w  e  w 1 e Gs  w  e  w 1 e - w  w ( Gs - 1 1 e ( Where... at Point A is = h1•w + h2•sat u = (h1 + h2 +h)•w ’  - u = (h1• w + h2•sat) - (h1 + h2 + h)•w = h2(sat - w) - h•w = h2•’ - h•w or  ’ h  ’  2    w   h 2 ’  i  w h h 2    “i” is the.../m3 4m Water Table B Clay sat = 17.2 kN/m3 5m C 1 - 30...) 0 (4)(d) = (4)(14.5) = 58 0 0 0 58 - 0 = 58 58 + (5)(sat) = (5)(17.2) = 144 (5)(w) = (5)(9.81) = 49.05.../m3 4m Water Table Clay sat = 17.2 kN/m B 3 5m C 1 - 31 Effective... Table Dry Sand sat = 15.2 kN/m3 B C Clay sat = 17.2 kN/m3... Water Table Dry Sand sat = 15.2 kN/m 3 B C Clay sat = 17.2 kN/m 3 5m...



Дата загрузки: 2017-01-16
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... A-2: Equilibrium constants of NiCl2(cr) + H2(g) U Ni(cr) + 2 HCl(g).........260 Table... galvanic cell: Pt H2(g, p = 1 bar) HCl(aq, aH + = 1, f H2 = 1) Fe(ClO4)2 (aq... more general cell reaction: Ox + 1 n H2(g) U Red + n H+ 2 (II.29) and of...]. O2 gaseous Al crystalline, cubic H2 gaseous Zn crystalline, hexagonal He... Equation (V.1): Ni(cr) + 2H+ U Ni2+ + H2(g), (V.2) and the corresponding standard entropy... Sections V.1.2 and V.2.1.3, as well as S mο (H2, g, 298.15 K) from the selected....6 g (NiSO4)/100g (H2O) which yields: m (sat) = (2.62 ± 0.05) mol·kg–1 (NiSO4...(cr)). The value of ln γ ± (sat) = – (3.326 ± 0.007) has been derived...]. The value of ln aw (sat) = – (0.0707 ± 0.0030) has been derived...ο (V.10) = – RT{2·[ln m (sat) + ln γ ± (sat)] + 7·ln aw (sat)} (V.12) which leads to... Equation (V.20) S mο (Ni2+) = ∆ f Smο (Ni2+) – S mο (H2, g) + S mο (Ni, cr) (V.20) S mο (Ni2+) = – (131...], [67RIZ/BID], [68CHA/FLE], [69MOR/SAT], [70HUE/SAT], [75MOS/FIT2], [76BER2], [78IWA... the pertinent auxiliary quantities for H2(g), H2O(g), CO(g) and CO2(g) listed... 10.78 pH, [Ni(II)] sat. at 35°C ? [49GAY/GAR] (b) (c) NiSO4... numerically equal to x (NiIII). β-NiOOH + ½ H2(g) U β-Ni(OH)2 (V.46) Figure V-12... difficult to interpret thermodynamically. γ-NiOOH + ½ H2(g) U α-Ni(OH)2 (V.51) Clearly the... case when the ratio γ [NiXn ( H2 O ) ] / γ [NiXn−1 ( H2 O ) ] = 1 over the whole range... measurements of the ratio p(H2S)/p(H2) after a hydrogen containing gas phase... of the reaction, NiCl2(cr)+H2(g) U Ni(cr) + 2 HCl(g) (A.3) using a static... account the wellknown equilibrium: H2S(g) U H2(g) + ½ S2(g). A. Discussion of selected references... A-2: Equilibrium constants of NiCl2(cr) + H2(g) U Ni(cr) + 2 HCl(g). T/K log10(Kp... following cell Ni | NiSO4 | KCl(sat.) | KCl(0.1 M), Hg2Cl2 | Hg. Murata’s data... cell: Ni | NiSO4 | KCl(sat.) | KCl(sat.), Hg2Cl2 | Hg. Their data have...([NiSO4]/m) -260 Ni|NiSO4|KCl (sat.)|KCl (sat.), Hg2Cl2|Hg E / mV (vs..., characterised by a certain ratio p(H2S)/p(H2), is in equilibrium with Ni3S2... correspond to the equilibrium: ½Ni9S8 + H2(g) U 3 2 Ni3S2 + H2S(g). In addition, the... the equilibrium reaction: NiCl2(cr) + H2(g) U Ni(cr) + 2 HCl(g) (A.9) in order... following reactions: Ni+H2S U ½Ni3S2+H2 –1 [∆ r Gm ]808K 673K = (– 75479.4 + 32... (T/K)) kJ·mol , 2Ni3S2+H2S U Ni6S5+H2 –1 [∆ r Gm ]798K 673K = (– 12761.2 – 7.7404... equilibrium concentrations of HBr(g) and H2(g) at equilibrium over solid samples..., T), Section V.4.1.4.2, C pο,m (Ni, cr, T), Section V.1.2, and C pο,m (H2, g) and C pο,m (HBr, g) [89COX/WAG], to... studying the equilibrium: NiCl2(cr) + H2(g) U Ni(cr) + 2 HCl(g) (A.10) and... formation equilibrium of hydrogen chloride: H2 + Cl2 U 2 HCl. A third law analysis... this textbook. The activity coefficient γ±(sat) has been calculated by a third... the activity of water, aW(sat), φ was extrapolated to the saturated... metallic nickel. T/K log10[p(H2S)/p(H2)], exp. log10[p(H2S)/p(H2)], calc. 853.15...) Ni(cr) + 2H+(sln, H2(sat)) U Ni2+(sln, H2(sat)) + H2(g) 7.068H2O(l) U 7.068H2O(sln) NiSO4...) + H2SO4·7.068H2O(l) U NiSO4(cr) + 7.068H2O(l) +H2(g) (A.24) at 303.15 K. The....07 ± 0.10) J·K–1·mol–1 (see Section V.1.2), C pο,m (H2, g, 298.15 K) = (28.836 ± 0.002... U NiSO4·6H2O + H2O(NiSO4(aq, sat)) was reported as (7.98 ± 0.21... Table V-6 and Figure V-11. [69MOR/SAT] The standard Gibbs energy of... constants for the reaction: H2S(g) U H2(g) + ½S2(g). [70ROT] Experimental levels of...-known gas phase equilibrium: H2S(g) U H2(g) + ½S2(g). Since this contribution is... constants for the reaction: H2S(g) U H2(g) + ½S2(g). In addition, the liquidus... the well-known equilibrium: H2S(g) U H2(g) + ½S2(g). Moreover, phase boundaries of... the well-known equilibrium H2S(g) U H2(g) + ½S2(g). Neither the initial compound... the gas phase equilibrium H2S(g) U H2(g) + 1/2 S2(g). The composition dependence of...(cr) – (411.260 ± 0.120) d (A.66) H2(g) + 2O2(g) + 6H2O + S(rh) U H2SO4·6H2O...: − φ ∑ mk k log10 aH2 O = (B.9) ln(10) ⋅ 1 M H2 O where φ is the osmotic coefficient...: log10 aH2 O = − 2 mNX φ ln(10) ⋅ 1 M H2 O (B.10) Alternatively, water activities can... in the galvanic cell: Pt H2(g, r) H+(r) 4+ + PuO 2+ 2 , Pu , H , H 2 O(l) Pt is: + + 4+ PuO... 2 O ⋅ aH+ (r )  . log10 K = log10   aPuO2+ ⋅ aH4 + ⋅ f H2 (r)  2   ο Since by definition of the SHE, f H2 (r) =1 and γ H+ (r) = 1 , log10 K ο = log10 K + log10g Pu..., R. S., Equilibrium in the reaction, NiO + H2 U Ni + H2O. The free energy..., R. H., Giauque, W. F., The equilibrium reaction NiCl2 + H2 U Ni + 2HCl. Ferromagnetism and the...: 111, 113, 114, 337. [69MOR/SAT] Moriyama, J., Sato, N., Asao, H., Kozuka, Z., Thermodynamic..., 340, 341, 343, 349. [70HUE/SAT] Huebner, J. S., Sato, M., The oxygen fugacity...] [74ARU], [75ARU] [63BOL/JAU] [69MOR/SAT] [70ASH] [77ASH/HAN] [77OSW/ASP...] [67RUD/DEV] [78KAR/HUB] [70HUE/SAT] [71PIE/HUG] [73HUL/DES] [50HUM...] [98ROG/KOZ], [2000ROG/KOZ] [69MOR/SAT], [79KEM/KAT] [55KRA/WAR] [76DAY...] [96STU/MOR2] [78IWA/FUJ] [69MOR/SAT] [71ARI/MOR] [64HAM/MOR] [65MOR.../CHI] [82SAR/COV] [70HUE/SAT], [97NOZ/KOB] [69MOR/SAT] [71DAS/DAS] [71LEB...



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0.07/5
... use the saturated unit weight sat . The dry unit weight d is... soil can be obtained as sat = W + Ww G +e W = s = s w V V 1+e w (1.41) Basic relations for... soil is saturated (Figure 1.34), sat = Ws + W w = 1 − n Gs V w +n w = Gs − n Gs − 1 w (1.44...), the saturated unit weight is sat Gs = w +e 1+e w = 9 81 2 68 + 0 8 = 18 97... = E2 E3 (3.98) a1 = b h2 (3.99) H= h1 h2 (3.100) Example 3.6 2 Refer to... 1 725 a1 = b 0 61 = =0 2 h2 3 05 H= 1 52 h1 = =0 5 h2 3 05 Stresses and displacements... given as +u = = h 1 + h2 where (4.1) (4.2) sat = effective stress u = pore water pressure = h2 w (4.3) w = unit weight of water Combining Eqs. (4.1)–(4.3) gives = − u = h 1 + h2 sath2 w = h1 + h2 where is the effective unit weight of soil = sat − w In... v = 0 5. 4.7 Redo Prob. 4.6 for point M; i.e., find h2 . Figure P4.2 Figure P4.3 Pore... equipotential lines along which h = h1 h2 h3 respectively. The slope along... x = 0 h = h1 2. at x = LA h = h2 However, h2 is unknown h1 > h2 . From the first...) into Eq. (5.113) gives C1 = h2 + h2 L L = h2 1 + A LB A LB (5.116) Thus, for flow through soil B, h=− h2 L x + h2 1 + A LB LB LA ≤ x ≤ LA + LB... through soil B So, q = kA h1 − h2 h2 A = kB A LA LB (5.118) where... h3 − h4 l3 b1 × 1 = k b3 × 1 = h2 − h 3 l2 b2 × 1 (5.122) 220 Permeability... Eq. (5.122), we get h1 − h2 = h2 − h3 = h3 − h4 = · · · = h = h Nd (5.123... z direction we can obtain 2 h z2 = 0 h2 + h4 − 2h0 z2 (5.142) Substitution... Numerical analysis of seepage. q2−0 = k h2 − h0 x z (5.147) q0−4 = k h0 − h4.... (5.145)–(5.148) into Eq. (5.149), 1 h + h2 + h3 + h4 4 1 If the point... q0−3 = k 233 h 0 − h3 z x 2 (5.151) h0 − h2 x z (5.152) q0−2 = k For continuity of... − h0 h0 − h3 − − h0 − h2 = 0 2 2 h1 h3 + + h2 − 2h0 = 0 2 2 or h0 = 1 h + 2h2... pile with hydraulic heads of h2 and h2 , respectively. For this condition... should follow Eq. (5.144); h1 + h2 + h3 + h4 − 4h0 = 0; or h i j+1 + h i−1 j + h i j−1 + h i+1 j − 4h... seepage is (length)(width)(thickness) sat = L L 1 sat = L2 sat . The hydrostatic force on... h2 L w . For equilibrium, F = h1 wL + L 2 However, h1 + L sin sat sin − h2 wL (5.166) = h2 + h, so h2 = h1 + L sin − h (5.167) Combining Eqs. (5.166) and (5.167), F = h1 wL + L 2 sat... − h1 + L sin − h wL or F = L2 sat − w sin + h wL = L2 sin + h wL... force of soil (5.168) where = sat − w . From Eq. (5.168) we can... in the soil mass: icr = w = sat − w = Gs w +e 1+e w − w = Gs − 1 1+e w So, icr = = w Gs... = qd = or q= H1 kz dz H2 k 2 H − H22 2 1 k H 2 − H22 2d 1 (5.181) Equation... conditions. Also note that if H2 = 0, the phreatic line would intersect... + = l sin2 s−l = 2ls sin2 − 2l2 sin2 H2 sin2 =0 (5.187) The solution to Eq. (5.187) is l = s − s2 − H2 sin2 (5.188) With about a 4–5% error.... (5.181)]. Substituting h1 for H1 h2 for H2 , and L for d in Eq... q= k 2 h − h22 2L 1 (5.192) where L = B + Hd − h2 cot 2 (5.193) 258 Permeability and... 2 (5.194) Combining Eqs. (5.191)–(5.193), h2 = B cot + Hd − 2 B cot 2 + Hd − h21... Figure 5.55, H1 = 25 m and H2 = 0; also, d (the horizontal distance between...), q = k l sin tan l= d cos − d2 cos2 − H2 sin2 Using Casagrande’s correction (Figure.... (5.198), we get h2 = 5 + 30 − 2 5 + 30 2 2 − h21 or h2 = 32 5 − 1056 25... · m = cot 2 2 Figure 5.56 Plot of h2 against h1 . 262 Permeability and... H1 = 3 m k1 = 4 × 10−3 mm/s d1 = 5 m H2 = 1 m k2 = 2 × 10−3 mm/s d2 = 5 m a b Draw... = m=0 m= = m=0 −M 2 C t − ta Mz 2dui sin exp M H H2 2dq Mz −M 2 C t − ta sin exp... Cv (2) = ¼Cv(1) H/2 Impervious (b) 1.0 Tv = Cv (1)t H2 Tv = 0.08 z /H 0.16 0.5 0.31 0.62... compression index. 336 Consolidation solution T = C t H2 For Uav = 50% T = 0 197. Hence...), and (6.98), we obtain 1/n 2 − a bn z¯ 2 1−n t = a H2 w k 1+e t = m H2 k w t = H2 C t (6.99) where m is the volume... be written in the form T = H2 C t (6.100) where T is the nondimensional... a nondimensional ratio R as R= bn T C a = Ts H2 (6.103) n−1 From Eqs. (6.99), (6.100.... 4. As defined in Eq. (6.103), R= C a H2 bn n−1 The term b is a complex.... (6.138), u= A1 H 2 C 2 or A1 = 2C u H2 (6.139) Substitution of this value... into Eq. (6.138) yields u = u 1− z2 H2 (6.140) Equation (6.140) shows a parabolic...) and (6.139), we obtain 2C u H2 H2 or C = t 2 u t = A1 = (6.141) Interpretation of... 2 t1 2. At time t1 < t < t3 , C = H2 t 2 u Note that / t H, and 3. Between time...−4 m/day = 1 807 m/year. T = C t 1 807 × t = = 0 2008t H2 6/2 2 0 2t (E6.1) Radial drainage: re... Table 7.5 and Eq. (7.64). solution sat clay = Gs w + wGs 1 + wGs = 19..., 3 LL = 68 PL = 29, and sat = 17 8 kN/m . Estimate the undrained... 5m G.W.T. 16 m Normally consolidated clay γsat = 19.1 kN/m3 Figure P7... of shallow foundations q G.W.T. ∆H1 ∆σ(1) Layer 1 ∆H2 ∆σ(2) 2 ∆σ(n) n Clay ∆Hn Figure 8.15 Calculation... for surcharge load removal, t, is t= T H2 C (8.58) where C is the coefficient... the desired goal. solution t= T H2 C or T = tC H2 528 Settlement of shallow... settlement. 3 Use Eq. (8.35) and sat = 17 5 kN/m Cc = 0 12, and....3 kN/m3 Sand γsat = 19.5 kN/m3 Clay γsat = 17.3 kN/m3...



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... type in Γ. The definition of “|= (s, h) sat (F, Γ.p)” splits the heap into two... ) F, s, h |= p1 ∗ p2 ∃h1 , h2 , s1 , s2 . (h = h1 ⊎ h2 ) ∧ (s = s1 ⊎ s2 ) ∧ (F, s1 , h1 |= p1 ) ∧ (F, s2 , h2 |= p2 ) F, s, h |= p1 −∗p2 ∀h1 . ((h1...(Γ). F |= s(x) : [[Γ(x)]] F, s, h |= Γ.p fvar(p) ⊆ dom(Γ) ∧ (F |= s : Γ) ∧ (F, s, h |= p) |= (s, h) sat (F, Γ.p) ∃h1 , h2 . h = h1 ⊎h2 ∧ dom(h1 ) = dom(F) ∧ (|= h1 : F) ∧ (F, s, h2 |= Γ.p) Fig. 9 Semantic... interpretation. Definition 3.6 (Partial correctness) Ψ |= {Γ.p} c {Γ′ .p′ } ∀F ≥ [[Ψ]], s, h. |= (s, h) sat (F, Γ.p) ⇒ ¬((s, h, c) −→∗ wrong) ∧ ∀s′ , h′ . (s, h, c) −→∗ (s′ , h′ , ǫ) ⇒ ∃F′ ≥ F. |= (s′ , h′ ) sat (F′ , Γ′ .p′ ) In the interpretation, it assumes... assume |= (s, h) sat (F, Γ.(e → e′ )). Therefore, for some h1 and h2 , we have h = h1 ⊎ h2 ∧ dom(h1 ) = dom(F) ∧ |= h1 : F ∧ F, s, h2 |= Γ.(e → e′ ) (10...(Γ)” (by F, s, h2 |= Γ.(e → e′ ) in (10)). The second subgoal is to show ∃F′ ≥ F. |= (s′ , h) sat (F′ , Γ′ .p′ ), where...), we only need to show F, s′ , h2 |= Γ′ .p′ , which is the same as (fvar(p′ ) ⊆ dom(Γ′ )) ∧ (F |= s′ : Γ′ ) ∧ (F, s′ , h2 |= p′ ) All the above can be...(e) {Γ[x ❀ wref τ ].emp} Pick F ≥ [[Ψ]], s, h and assume |= (s, h) sat (F, Γ.emp). Therefore, we have dom... second subgoal is to show ∃F′ ≥ F. |= (s′ , h′ ) sat (F′ , Γ′ .emp), where s = s[x ❀ ℓ], h′ = h ⊎ {ℓ ❀ s(e)}, and Γ′ = Γ[x ❀ wref τ ]. ′ Let... us |= (s′ , h′ ) sat (F′ , Γ′ .emp). ✷ Lemma 3.3 (Locality) If s = s1 ⊎s2 , h = h1 ⊎h2 , ¬(s1 , h1... exist s′1 and h′1 such that s′ = s′1 ⊎s2 , h′ = h′1 ⊎h2 , and (s1 , h1 , c) −→ (s′1 , h′1 , c′ ). §4 Concurrency In... for commands. Definition 4.1 (Partial correctness) ΓG .I, Ψ |=c {Γ.p} c {Γ′ .p′ } ∀F ≥ [[Ψ]], s, h. |= (s, h) sat (F, (ΓG .I) ∗ (Γ.p)) ⇒ ¬((s, h, c) −→∗ wrong) ∧ ∀s′ , h′ . (s, h, c) −→∗ (s′ , h′ , ǫ) ⇒ ∃F′ ≥ F. |= (s′ , h′ ) sat (F′ , (ΓG .I) ∗ (Γ′ .p′ )) Theorem 4.1 (Soundness) If ΓG .I, Ψ ⊢c {Γ.p} c {Γ′ .p′ }, then... separation logic 25 2. if c = ǫ, then |= (s, h) sat (F, (ΓG .I) ∗ (Γ.p)). 3. for all s′ , h′ , and c′ , if (s, h, c) −→ (s′ , h′ , c′ ), then... F′ ≥ F and ∀j ≤ k. ΓG .I, F′ |=jc (s′ , h′ , c′ ) : Γ.p; 4. there exist h1 , h2 , s1 and s2 such that... ⊎ h2 , s = s1 ⊎ s2 , and a. |= (s1 , h1 ) sat (F, ΓG .I); b. for all h′1 , s′1 and F′ , if h′1 ⊥h2 , s′1 ⊥s2 , F′ ≥ F, and |= (s′1 , h′1 ) sat (F′ , ΓG .I), then, ∀j ≤ k. ΓG .I, F′ |=jc (s′1 ⊎ s2 , h′1 ⊎ h2 , c) : Γ.p. We... there exists F′ such that F′ ≥ F and |= (s′ , h′ ) sat (F′ , (ΓG .I) ∗ (Γ.p)). Proof. By induction over the... of execution steps. ✷ Definition 4.3 ΓG .I, Ψ |=sc {Γ.p} c {Γ′ .p′ } ∀F ≥ [[Ψ]], s, h. |= (s, h) sat (F, (ΓG .I) ∗ (Γ.p)) ⇒ (ΓG .I, F |=c (s, h, c) : Γ′ .p′ ) Lemma 4.2 If ΓG .I, Ψ |=sc {Γ.p} c {Γ′ .p′ }, then ΓG .I, Ψ |=c {Γ.p} c {Γ′ .p′ }. Proof... {Γ′2 .p′2 }. For all s, h and F, if F ≥ [[Ψ]] and |= (s, h) sat (F, ΓG .I ∗ Γ.p), show ΓG .I, F |=c (s, h, c) : Γ′ .p′ . We know there exist... and h2 such that s = s0 ⊎ s1 ⊎ s2 , h = h0 ⊎ h1 ⊎ h2 , |= (s0 , h0 ) sat (F, ΓG .I), F, s1 , h1 |= Γ1 .p1 , and F, s2 , h2 |= Γ2 .p2 . Gang... , c1 ) : Γ′1 .p′1 ; and (2) ΓG .I, F |=c (s0 ⊎s2 , h0 ⊎h2 , c2 ) : Γ′2 .p′2 . Our final goal ΓG .I, F |=c (s, h, c) : Γ′ .p′ can... ⊎ s1 ⊎ s2 , h = h0 ⊎ h1 ⊎ h2 , |= (s0 , h0 ) sat (F, ΓG .I), ΓG .I, F |=kc (s0 ⊎s1 , h0... , and ΓG .I, F |=kc (s0 ⊎s2 , h0 ⊎h2 , c2 ) : Γ2 .p2 , then we have... ⊎h1 ) sat (F, (ΓG .I) ∗ (Γ1 .p1 )). Similarly we can prove |= (s0 ⊎s2 , h0 ⊎h2 ) sat (F, (ΓG .I) ∗ (Γ2 .p2 )). Since |= (s0 , h0 ) sat (F, ΓG .I) and I is precise, we have |= (s, h) sat... exist s′′ , and h′′ such that s′ = s′′⊎s2 , h′ = h′′⊎h2 , and (s0 ⊎s1 , h0 ⊎h1... that s′′ = s′0 ⊎ s′1 , h′′ = h′0 ⊎ h′1 , and |= (s′0 , h′0 ) sat (F′ , ΓG .I). Then, by ΓG .I, F |=i+1 (s0 ⊎s2 , h0 ⊎h2 , c2 ) : Γ2 .p2 and... know for all j ≤ i, ΓG .I, F′ |=jc (s′0 ⊎s2 , h′0 ⊎h2 , c2 ) : Γ2 .p2 . By the induction...



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... of g1 modulo {g2 , . . . , gk }, h2 be obtained by reduction of... f . If f is any binomial, let Sat(f ) be the binomial obtained from f by saturation, i.e., Sat(f ) is a saturated binomial such that f /Sat(f ) is a monomial. Let Bsat be ... f1 , . . . , fr , let m1 := φ−1 (Sat(f1 )), . . ., mr := φ−1 (Sat(fr )); we denote by... generated by m1 , . . . , mr . By Sat(I) we denote the saturation of... I with respect to the indeterminates, i.e., Sat(I) = {f ∈ K[x1 , . . . , xn ] : ∃m, a monomial, s.t. mf ∈ I} . Finally... . Then: 1. φ(b1 − b2 ) = Sat(σ(φ(b1 ), φ(b2 ))); 2. φ(b1 + b2 ) = Sat(σ(φ(b1 ), φ(−b2 ))). Let... ) ∈ Sat(I), then φ(b1 − b2 ), φ(b1 + b2 ) ∈ Sat(I); 4. if φ(b1 ) ∈ Sat(I) and λ ∈ Z, then φ(λb1 ) ∈ Sat... that φ(b1 − b2 ) = Sat(xp φ(b1 − b2 )) = Sat(σ(φ(b1 ), φ(b2 ))). The second..., by definition of σ, σ(φ(b1 ), φ(b2 )) ∈ Sat(I) and hence the statement follows... , . . . , fr . If b ∈ M (f1 , . . . , fr ), then φ(b) ∈ Sat(I). Proof. By definition, M (f1 , . . . , fr... ) and the corresponding φ(mi )(= Sat(fi )) are in Sat(I) (i = 1, . . . , r). If b ∈ M then b is... 4.4. If I is a binomial ideal, then Sat(I) = (1) and is a binomial ideal. Proof... G are pure binomials. Hence Sat(I) = (1). The ideal Sat(I) can be computed by the formula: Sat(I) = K[x1 , . . . , xn ] ∩ (I + (t · x1 · · · xn − 1)) , and from this it follows that Sat(I) is a pure binomial ideal. Proposition... , . . . , fr ) is a binomial ideal and f ∈ Sat(I) is a saturated binomial, then there... ∈ M (f1 , . . . , fr ) such that φ(pi ) = Sat(gi ) for i = 1, . . . , k. To prove this... computation of the S-polynomial S(a, b). Since Sat(f1 ), . . . , Sat(fr ) are in φ(M (f1... to prove that if h1 , h2 ∈ I are two binomials whose saturation... also Sat(S(h1 , h2 )) is in φ(M (f1 , . . . , fr )). Let then h1 , h2 ∈ I and let b1 , b2 ∈ M (f1 , . . . , fr ) be such that φ(b1 ) = Sat(h1 ) and φ(b2 ) = − + − Sat(h2 ). Let b1 = b+ 1 − b1... h2 with −b2 and −h2 ). Then Sat(S(h1 , h2 )) = Sat(σ(φ(b1 ), φ(b2 ))). Lemma 4.2 gives that Sat(S(h1 , h2 )) = φ(b1... the claim follows. Let now f ∈ Sat(I) be a saturated binomial; then there... − xts+1 gis+1 we deduce that Sat(us ) is in φ(M (f1 , . . . , fr... that Ltτ (v) = Ltτ (g), u = v − g and Sat(u), Sat(g) ∈ φ(M (f1 , . . . , fr )). Then Sat(v) ∈ φ(M (f1 , . . . , fr )). Let... , . . . , fr ) be such that φ(b) = Sat(u) and φ(p) = + − + − Sat(g). Then u = xb +c − xb +c , g = xp +d − xp... +d − xb +c . Consider p + b: p + b = (p+ + b+ ) − (p− + b− ) = (p+ + b+ + c) − (p− + b− + c) = (p+ + d) − (b− + c). This shows that Sat(v) = φ(p + b). As a consequence of Proposition 4.4, we... have a Gr¨ obner basis for Sat(I) formed by binomials; moreover it... on, a Gr¨ obner basis for Sat(I) will always 56 G. BOFFI AND... of all saturated binomials of Sat(I). In particular, φ(M (f1 , . . . , fr )) contains... the Gr¨ obner bases of Sat(I). Suppose that I = (f1 , . . . , fr ) = (h1... of all saturated binomials of Sat(I). The injectivity of φ gives M (f1... , . . . , hs ) (since every saturated binomial f ∈ Sat(I) is equal to both φ(b) ∈ M (f1...¨ obner bases of the ideal Sat(I). More precisely, if <τ is a term... φ(G) is a Gr¨ obner basis for Sat(I) w.r.t. <τ , and conversely. Furthermore, in this...¨ obner bases for the ideal Sat(I). Proof. Let <τ be a term order... φ(G) is a Gr¨ obner basis for Sat(I) w.r.t. <τ . From Lemma 4.7, it is enough... that if f ∈ Sat(I) is a binomial, then Ltτ (f ) ∈ Ltτ (φ(G)). Let h ∈ Sat(I) be the... H is a Gr¨ obner basis for Sat(I), then the same kind of... (reduced) Gr¨ obner bases of Sat(I). We end this section by... ]. Proposition 4.9. For every binomial ideal I, Sat(I) equals (φ(MI )), the ideal generated.... By Proposition 4.3, (φ(MI )) ⊆ Sat(I). By Proposition 4.6, (φ(MI )) ⊇ Sat(I). Proposition 4.10. For every... Proposition 4.9 says that (φ(MI )) = Sat(I). Hence (φ(M )) = Sat(I). Corollary 4.12. For every binomial ideal I, MI = MSat(I) . Proof. Let J := Sat(I). Proposition 4.9 says that J = (φ(MI )). But... φω = Id because Propositon 4.9 implies (φ(MJ )) = Sat(J), and Sat(J) = J by assumption. ωχ = Id because... for MI , and hence for Sat(I): Let Bi ⊆ A (i ∈ N) be any family...) lex Gr¨ obner basis for Sat(I). The module MI is generated... 3: The Gr¨ obner basis for Sat(I) C3 := C(1, −1) = {(u, v) ∈ A : u ≥ 1, v ≤ −4/9u − 5/9}. Choose B2 := {(2, −1), (3, −1)}. The... reduced Gr¨ obner basis for Sat(I) (w.r.t. the lexicographic term order in...



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...:E / 1127. 4 (9+12) 1128. = B® (4+4) 1129. + (5+5) 1130. h2 (4+8) B)Vc2®d2 - 202 – Originály... : : : / /t: 01r : :/ /E:E E:E E:E E:E / /ð ð:ð ð:ð ð:ð ð:/ /: : 67F : / / : :P: :/ /: :R()P : / / : 23D : :/ /: : :r: / / : 89c : :/ /E:E E:E E:E E:E / 1131. h2 2 sol. (6+9) 1132. h3 B)Re6®f6... PONGRÁCZ 10“ oddelenie exo-sat, C 13. 8. 2002 Správa usporiadate...á 27, 010 07 Žilina, Slovakia) SAT – ak sa nedá zrušiť vo... očakávania. Predovšetkým v tom, že SAT ukázal nové priestory pre... známe exokamene a syntézy SAT s inými exopodmienkami. 67 skladieb... komponovaných v základnom sate. Ostrý sat a sat x/y zrejme čakajú na svojich prieskumn...é originály (syntézy, ostrý sat a sat x/y). Toto rozhodnutie odzrkadľuje snahu... aequo 1143 Juraj Lörinc, Slovensko sat, na poliach a1, a5, a6.../ /t:P45T :C()P/ /E01rE E01REFE:E E:E / 1144. 23 sat (6+10) 1145. 2 text (9+14) 1146. 34 sat (5+13) 4.Vg4+ Jf2 (4.Vg3? Jf2... 1145 Juraj Lörinc, Slovensko sat, tátoš b1, h1 a h2, ťava b6 a e6...ívov, ktoré ponúka len sat. Dodávam, že mnohé ešte čakaj...á ako šachujúci kameň, B) Fh8®h2, C) Fh8®a1, zvodnosť 1.a8J? hroz.../ /:G=?Gn01r =?G / /T: : : :/ /E:E E:E E:E E:E / 1147. 2 text (4+7) 1148. 11 sat (11+8) 1149. s2 text (5+14.... 15 sat (5+9) 1. čestné uznanie 1148 Ľuboš Kekely, Slovensko 1.Sh6 (1.Vxc5? h2 2.Vxa5...! 8.Vb3 Dd4!) 1.- Sa3 2.Vg5+ (2.d4? h2 3.dxc5 b3! 4.Vg5+ Ka2! 5.c6... 6.Vg5+ Sc1 7.Vb5 Sa3 8.Vxb4 h2 9.Vb3 h1D 10.Vxa3 De1...!5.Vf4+ Sc1 6.Vf6 b2! 7.Vb6 h2 8.Vb3 h1D 9.Ka2 Ka1! 10...é uznanie 1149 Juraj Lörinc, Slovensko sat, na poliach d7, e7 a f8...é uznanie 1151 Juraj Lörinc, Slovensko sat, na poli g7 je lion... text (8+10) 1152. 7 sat (4+12) 1153. h2 sat (3+10) 4 sol. 1154. h2 sat (7+16) 2 sol...é, ale pozorné skúmanie presved čí, že sat aj v pomocnom mate má originálne...á zmienka 1156 Karol Mlynka, Slovensko sat, na b5 je prešporsk... E:E E:E E:Et/ 1155. + sat (7+6) 1156. 2 text (7+7) 1157. h= =9 sat (6+5) 1158. s2 sat (6+6) hrozí 2.c6X, 1.- Ja... TT Pongrácz 10 – oddelenie sat – 1. skupina môžete posielať do 30...



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... /ð ð:ð ð:ð ð:ð ð:/ /: : ()P 89c / / : : : :/ /: :P: 45t / / : : : :/ /: : 01R : / / : : : :/ /E:E E:E E01rE E:E / 1273. h2 2 sol. (3+3) 1266. 6 (8+8) Michal Hlinka Ko...: / / : 23D 45T :/ /E:E E:E E:E E67FEF/ 1274. h2 2 sol. (5+4) 1275. h2 2 sol. (3+11) - 227 Najbohat... E:E E:E E:E / /ðTð:ð ð:ðRð:ð ð:/ /: : : : / / : 67FC: :/ /: ()pf()p : / / : : : :/ /: : : :F/ / : : : 01r/ /E:E E:E E:E E:E / 1276. h2 2 sol. (5+9) 1277. h2 text (6+11) Attila Benedek... Vráble Snežnica 1290. 2 sat (3+9) Ion Murarasu Rumunsko 1291. 2 text...:E E:E / /ð ð:ð ð:ð ð:ð ð:/ /: : : : / / : : : :/ /: : ()¹ : / /¹: 01R : :/ /: : : : / / : : : :/ /E:E E:E E:E E:Er/ 1293. 13 sat (8+12) 1294. h2 text (1+1+2) Béla Majoros Zoran... Maďarsko Macedónsko 1295. h2 text (5+7) Milan Ondruš Žilina /ð ð&ŽSðCð:ð ð:ð ð:/ /:p()p :S: / / ()Pf01rt... nad Váhom h2 sat (6+15) 2 sol. sh10 (4+4) /ð ð:ð ð67fðrð67fð ð:/ /: : : : / / :P:R: :/ /: : 89C : / / :P:p:P:/ /:P: :P: / / : : ()p :/ /E:E E:E E:E E:E / h= =4 sat (6+6) - 229 Imants... / 1300. s6 max. (4+3) 1301. s8 sat circe (9+6) 1302. sh19 text (5+2) 1303...é motívy šité na mieru sat s diagonálno-ortogonálnym echom... cyklus Sat, sat, sat. Mat, mat, mat. Pat, pat, pat. Pat, pat, pat? Sat, sat, sat? Mat, mat, mat? Mat, mat, mat! Pat, pat, pat! Sat, sat, sat! Pat-sat : mat. Sat-mat : pat? MAT-PAT : sat! (Pat a mat? – SAT!) Medzin...



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... of the Saturated Phases of H2 O from 173.15 K to 473... The Saturator Inlet/Outlet Heater H2 is a resistive heating element which... … C PP Mv PPM w %RH S A T U R p si SAT U R …C TES T p si TES T …C FL OW... … C PP Mv PPM w %RH S A T U R p si SAT U R …C TES T p si TES T …C FL OW... W PT … C PP Mv PPM w %RH SAT U R p s i SAT U R …C TES T p si TE ST …C FL... W PT … C PP Mv PPM w %RH SAT U R p s i SAT U R …C TES T p si TE ST …C FL... … C PP Mv PPM w %RH S A T U R p si SAT U R …C TES T p si TES T …C FL OW... … C PP Mv PPM w %RH S A T U R p si SAT U R …C TES T p si TES T …C FL OW... Pressure Zero Low Range Sat Hi Range Sat External Test Supply _____... F2 FB1 FD1,2 G1 H1 H2 HLS1 KB LCD-CON LCD... RT1 RTD1 RTD2 RV1 S1 SAT SL1,2 SM-1,2 SMD-1,2 SOL1-4 SOL5... AF2-X0-03-947-531-C SAT-3900 MCRSW2 PK243B1A-SG18 RD... APPROVED GMF A RENAMED METH. TO SAT. FLUID & RTD5 TO TR5. 9/28... DWN REV DESCRIPTION GMF A ADDED H2 SAT HEATER BKS B CLARIS TO MINICAD... SOLENOID 11 (-) 12 SSR5 13 H2 SAT IN/OUT HEATER TB1-1 (240... C RV1 TR1 REG G1 TR2 SAT FILL SOL1 V1 GAS INPUT... C RV1 TR1 REG G1 TR2 SAT FILL SOL1 TR4 TR3 V1... EXPANSION TANK FLUID FILL PORT SAT FLUID RTD 1 FILL PORT RT1... %RH = 10.39 Sat Pressure = 34.73 psiA Sat Temperature = -0.01 °C Test... = 1771 %RH = 11.44 Sat Pressure = 131.5 psiA Sat Temperature = 10 °C Test...



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...) thermistor or Supply Air Temperature (SAT) thermistor input, SA Bus network... sources. Supply Air Temperature (SAT) The SAT displays the current UCB thermistor... Off Due to Gas Valve H2 with no GV2 for >=6 minutes... Air Temp Cooling SAT < Excessive SAT Cooling Sp AND SAT Limit for Cooling... Valve Failure GV2 on without H2 for >= 5 seconds HS3 Gas Valve... Air Temp Heating SAT > Excessive SAT Heating SP AND SAT Air Temp Limit... Sequencer = DAT Control and SAT Unreliable or SAT Unrealiable and MA Sequence...) (73.0 F) (Space Temperature Input) OprST SAT (60.7 F) RAT (73.0 F) (Return Air... (Staged Cooling Command) Off H2 Off (UCB H2 24 VAC output status...(s) for Parameter To Be Displayed SAT (60.7 F) (S A Temp Thermistor input) RAT...) SATTempHydHt-En No (Hyd Heat SAT Tempering Enabled) SATTempHydHt-Sp Econ 40F (Hyd Heat SAT Tempering Sp) Econ-En Yes... F ClgUnocc-Sp 80 F CVHtgOcc-SP SAT (CV - UnOcc Cooling Setpoint) ( CV... Speed (UnitOpMode) APS Status DctPrs SAT Off (APS input status) 1.50"/w (DuctPres 0-5vdc input) 60.7 F (UCB SAT thermistor input) FanOverload Normal (FanOvrInptStatus... (Cooling Status) #ClgStgs SAT 2 (# of cooling stages) 60.7 F (SAT thermistor input) StgClgCmd... SATCoolLimit-Sp Yes (Enable SAT Limit) 50 F (SAT Limit SetPt) EconLoad-En... ASCD Time) H2RunTim .0 hr (accum H2 RunTime) H3 Stage3 Off-Idle... H3-S Htg 0 min (RemainMinRunTime) H1ASCDTmr H2 Details (Continued) (Heating Stage Status... Stg Heat output status) H1OnTmr H2-S Stage2 Off-Idle Off-Idle... F (Hyd H1 SAT SetPt) HydH2SA-Sp 150 F (Hyd H2 SAT SetPt) SATTempHydHt-En...-En Yes (Enable SAT Limit) SATCoolLimit-Sp 50 F (SAT Limit SetPt) SATHtgLimit... Input) OAQ SAT 477ppm (OAQ 0-10vdc Input) 60.7 F (UCB SAT ThermistorInput) RAT... (1st Heat 24 VAC output) H2 Off (2nd+ Heat 24 VAC... 29563 307 #N/A 0 #N/A 0 0 Supply Air Temperature SAT 29564 158 X10 0 Signed 0 0 Duct... erDegOff 29589 24 X10 0 Signed 0 1 SAT Limit for Clg Ena SATCoolLimitEn 29590 98 #N/A 0 #N/A 0 1 SAT Limit for Cooling Sp SATCoolLimitSp... #N/A 0 #N/A 0 1 Heating Status Htg-S 29708 347 #N/A 0 #N/A 0 0 SAT Limit for Htg Enable SATHtgLimitEn 29709 106 #N/A 0 #N/A 0 1 SAT Limit for Htg Sp SATHtgLimitSp... X10 0 Unsigned 0 1 Heating Stage 2 Command H2 29733 361 #N/A 0 #N/A 0 0 Heating Stage 2 Status H2-S 29736 362 #N/A 0 #N/A 0 0 Heating Stage 2 Runtime ... 29742 115 #N/A 0 #N/A 0 1 HW Htg Stage 1 SAT Sp HydH1SA-Sp 29743 51 X10 0 Signed 0 1 HW Htg Stage 2 SAT Sp HydH2SA-Sp 29744 52 X10 0 Signed 0 1 HW Heat SAT Tempring Ena SATTempHyd Ht-En 29745 116 #N/A 0 #N/A 0 1 HW Heat SAT Tempering Sp SATTempHyd Ht-Sp... BI 37 Heating Stage 2 Status H2-S SE RTU Heating Stage 1 Status... BO 11 Heating Stage 2 Command H2-C Off/On BO 12 Heating... Offst THVDGOFF 5 %RH ADI 33 SAT Limit for Clg Ena SATCLMEN Yes No/Yes ADF 46 SAT Limit for Cooling Sp SATCLMSP... Heating Control Status ADI 66 SAT Limit for Htg Enable SATHL... Yes No/Yes ADF 116 SAT Limit for Htg Sp SATHL... (2) ADF 130 Heating Stage 2 Runtime H2-RT Heating Stage (3) ADF 131... ADF 132 HW Htg Stage 1 SAT Sp HWH1SASP 120 deg F ADF 133 HW Htg Stage 2 SAT Sp HWH2SASP 150 deg F ADI 80 HW Heat SAT Tempring Ena SATEMPEN No No/Yes ADF 134 HW Heat SAT Tempering Sp SATEMPSP 40 deg...



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... the coed school on the SAT 1 mathematical reasoning section. Finally, no..., and Validity and Reliability.................... 28 SAT 1: Reasoning Test................................................................... 28 High School... of the Difference Between the SAT Mathematical Reasoning Mean Scores of... Analysis of Covariance of the SAT Verbal and Mathematical Reasoning Scores... and Standard Deviations of the SAT Verbal and Mathematical Reasoning Scores... 4 T Test of Means Comparing the SAT Verbal and Mathematical Reasoning Mean... 7 Analysis of Covariance of the SAT Verbal and Mathematical Reasoning Scores... 8 T Test of Means Comparing the SAT Verbal and Mathematical Reasoning Scores... as indicated by the SAT 1: Reasoning Test (SAT) verbal reasoning scores was... abilities as indicated by the SAT mathematical reasoning scores was the... schools score higher on the SAT mathematical reasoning sections than their... coed school on the SAT 1 verbal reasoning section. H2 There is 110... the coed school on the SAT 1 mathematical reasoning section. H3 There... had an effect on either SAT verbal reasoning or mathematical reasoning... classroom sex mix, SAT verbal ability scores, SAT mathematical ability scores, and..., Furthermore, since students take the SAT exam at different points in their high school careers, SAT 1 scores from the October, November... school students to take their SAT 1 exams. However, this further limited... of the study, the SAT I: Reasoning Test (SAT) verbal and mathematical ability.... The specific data collected were SAT scores, High School Placement Test.... Application, and Validity and Reliability SAT I: Reasoning Test The first part... verbal and mathematical reasoning. The SAT I: Reasoning Test was the instrument... for several reasons. First, the SAT measures the extent of development...). 28 The question of the SAT's validity has been exhaustively researched..., 1985). The studies confirmed that SAT results were correlated with college..., and Cohn (1985) claimed the SAT was the "best documented" .instrument... has not supported accusations that SAT results "underpredict" college grade averages... later statistical comparisons of the SAT results. The test publisher's own... relationship between HSPT scores and SAT scores. Their report pointed out... students in general," but the SAT normative populations "represent very elite.... 35 The collection of the SAT verbal and mathematical reasoning scores... calculated for the variables: SAT verbal score, SAT mathematical score, HSPT verbal... coed school on the SAT 1 verbal reasoning section, SAT verbal reasoning scores... coed school on the SAT 1 mathematical reasoning section, SAT mathematical reasoning scores... Hypothesis Data Methodology Statistic Hi SAT verbal reasoning scores and HSPT... for means, analysis of covariance H2 SAT mathematical reasoning scores and HSPT... the difference between the SAT l:Reasoning Test (SAT) verbal reasoning mean scores... of the difference between the SAT mathematical reasoning mean scores of... school on the SAT 1 verbal reasoning section. 40 H2 There was no.... A nalpkM the„D ifemc.£Betweeii _the SAT Verbal Reasoning Mean Scores of... significant difference between the mean SAT verbal scores of the comparison...-Pevialions of the_SAI_L^RgasQDiiig Test (SAT) Verbal and Mathematical Reasoning Scores... 4 T Test, of Means Comparing the.SAT ..Verbal and_Mathematicil^£aMflingMean_Scores fl£SfinloxJ3irls... of the Difference Between ..the SAT Mathematical Reasoning Mean Scores Jnji... the coed school on the SAT 1 mathematical reasoning section, the means... the t test that compared the SAT mathematical reasoning mean scores of... analysis of covariance of the SAT verbal reasoning produced an f-value... analysis of covariance of the SAT mathematical reasoning produced a much lower... 7 Analysis of Covariance of the SAT Verbal and Mathematical Reasoning Scores... Covariate (n = 247) Variable n Observed M Adjusted M SAT Verbal Reasoning Girls School 123... the mean differences for the SAT verbal and mathematical reasoning scores... 1 tests also yielded t-values (SAT verbal, p = .443; SAT mathematical, p = .556) that were... in ANCOYA Sample (n = 247) Variable M m n SAT Verbal Girls School 513.17... Variable M difference t-value df p-value SAT Verbal 7.77 0.77 245 .443... the coed school on the SAT 1 mathematical reasoning section. Finally, no... outcome areas: cognitive and affective. SAT verbal and mathematical reasoning scores... schools score higher on the SAT mathematical reasoning section than their... possibility that those without recorded SAT scores in each school differ... instead of the SAT. Other students took the SAT later in their... the coed school on the SAT 1 mathematical reasoning section. TMs failure... the coed school on the SAT verbal reasoning section. 2. No significant... the coed school on the SAT mathematical reasoning section. 3. No significant... on the PSAT, ACT and SAT (Research Bulletin #134). (1987). Bensenville... B IIsiafcifiglLifcdiiQflLHag^ Scores HSPT Composite SAT Verbal SAT Math 99 98 98 95... on the PSAT, ACT, and SAT", Research Bulletin #134, 1987, p. 1. Copyright...