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9 8 3 1 7 6 1 1 5 3 1 4 3 2 1 0 / @ 9 7 6 5 = ;. This has determinant (ad bc) N(ld mc)Since ad bc = 1 (mod N), and c,d are coprime, we can find l,m such that (ad bc) N(ld mc) = 1This gives a lifting of the original matrix to SL2(Z) (d) Give an example of an integer N and an element of GL2(Z/NZ) which is not in the image of GL2(Z) Solution Any element of GL2(Z) has determinant 1, so its image modulo N has deter. 9 b 0 a 1 @ 3 2 1 4 0 1 3 2 1 4 2 9 1.

Z l , a Z g b a Z p b b i j. 9 8 7 6 5 4 3 j > = i > 8 ?. Calculate the probability you entered from the ztable of p(z 15) The ztable probability runs from 0 to z and z to 0, so we lookup our value From the table below, we find our value of.

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23/01/17 · 2575 < Z < 2575 Total probability =1 then, 1095=05 divide by 2, 05/2 = 0025 From normal distribution table, we found that P(Z < 0025) =2575 Therefore P(X < Z < X) =095 = P(2575 < X < 2575). _____ *Please check current. 2 * 0 % / , !.

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MODULAR FORMS EXAMPLE SHEET 4 1a Show that the group SL 2(Z=pZ) (pprime) has order p(p2 1) The rows of an element of GL 2(Z=pZ) are a pair of linearly independent vectors v 1;v 2 in (Z=pZ)2There are p2 1 choices for v 1, and for any xed v 1, pvectors v 2 that are scalar multiples of v 2Thus there are p2 pvectors v 2 such that v 1;v 2 is linearly independent The matrix with rows v. 9 Повышение точности удара по мячу > e y l h h q l h m e m q r b l h q g h k m ^ Z j Z i h f y q m, g _ h o h ^ b f h. ^ j m b o k e m q Z y o h g Z k d h j _ _ f _ r Z _ l q _ f i h f h Z _ l \ h k i j b y – –.

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