... ( TPOL ) + 0.138 2 ) 0.216 ( TPOL ) + 0.407 1 ) 0.119 ( TPOL ) - 0.111 2 ) 0.119 ( TPOL ) + 0.060 1 ) 0.014 ( TPOL ) + 0.005 2 ) 0.014 ( TPOL ) + 0.079 1 ) 0.386 ( TPOL ) - 1.702 2 ) 0.399 ( TPOL ) - 1.502 1 ) 0.099 ( TPOL ) - 0.490 2 ...
... TPOL $ $ CALC TCIG = % GL ( 6,6 ) TPOL = % GL ( 6,1 ) $ The variates TCIG and TPOL each take values 1 to 6 , but are of length 36 . Cell sample sizes are spread fairly evenly from 2 to 11 , with two larger values ( 14 and 17 ) . One ...
... TPol qk roots +n of generates q are on an + the extreme unit circle; ray equivalently, of TPoln if and only if qn = 0 and all the q(z) is of the form q(z) = c Å 1z ã , (1.1.5) n∏ (z − eitk) − e−itk k=1 for some c > 0 and tk ∈ R, k ...
... TPOL ( 3 ) 9 0.9983 0.7766 TPOL ( 4 ) 10 1.251 0.8010 TPOL ( 5 ) 11 2.353 0.9525 TPOL ( 6 ) scale parameter 1.000 $ print % rs $ 0.9595 -0.9809 0.5280 -0.0041 -0.4857 0.2948 -0.4437 -0.0045 -0.0068 -0.0028 -0.0032 -0.0032 -0.0030 ...
... TPOL . Finally , since TPOL , unlike FPOL , possesses hydrogen atoms at the " e " positions ( refer to Fig . 20.20 ) , the reaction proceeds to Stage 3 , in which the tip of one fused aromatic system undergoes an electrophilic aromatic ...
... TPOL. Sanders, Baker, and Turner (2012) compared TPOL (n = 60) with an Internet-use-as-usual control condition (n = 56) for parents of children aged 2–9 years. Parents participating in the TPOL condition reported significantly fewer ...
... Tpol pol Opal Mountain 8 TR Tpol 5 10 ' 5 ' Tbs BLACKWATER A Black Mountain Tbs Tbs PTi Tbm 3600 3400 3200 Tbm BLACK MTN FAULT 2 க்க 1 2800 3000 2600 2200 1 0 10 ' FAULT Tbs SUPERIOR pTi 3 Tam Tbm SUPERIOR LAKE VALLEY Qu 117 ° 00 ...
... Tpol Tbs BLACKWATER apot 3000 Tp Opal Mountain Tpol 5 Tbm FAULT Tbs A Black Mountain Tbs Tbs 3600 3400 3200 Tbm BLACK MTN FAULT HARPER LAKE VALLEY HARPER LAKE 15 ' 2 pTi 2800 3000 2600 2200 10 ' Tbs Tbm SUPERIOR SUPERIOR LAKE VALLEY Qu ...
... Tpol 1 / Tk + Dq2 ( 6.165 ) characterized by the relaxation times τκ = bk Tpol 2/3 – bk ( 6.166 ) where we have used expression ( 6.150 ) for the ratio Tpol / Te . From ( 6.159 ) , we can calculate the expression for the eigenmodes To ...
... Tpol 200 ° C , S - type polymer ( d = 43 A w ; 22 A s ; halo 4.4 Å ) = d ) Tpol 250 ° C , S - type polymer ( d 43.5 Aw ; 22.2 Å s ; halo 4.4 Å ) a ) Tpol polymer [ 156 ] 140 C I = 140 ° C , N - type [ 157 ] b ) Tpol polymer = 170 ° C ...