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MCP=1.9 kv

MCP=1.9 kv

MCP=1.9 kv. Mcp 1900 again. Mcp 1925. 1950. 1850. 1875. 1875 with 0-1 volt scale. 1875v, 0-1V, 700MHz filter. SI area 17.1.

By natalie-lucas
(148 views)


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0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0

0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0

Create the CWCT (table of counts of all Class value bitmaps ANDed with all Feature Value bitmaps) for InfoGain , Correlation. Chi Square, etc. CWCT has # s needed for IG, Corr., Chi 2 in attribute selection , DTI, etc. S 1,j = 0 0 0 0 3 0 0 3 0 3.

By genglish (0 views)

1 0 0 –1 1 0 –1 0 1

1 0 0 –1 1 0 –1 0 1

n 1 2 3 4 5 6 7 8 9. A n 1 2 7 42 429 7436 218348 10850216 911835460. = 2  3  7. = 3  11  13. = 2 2  11  13 2. = 2 2  13 2  17  19. = 2 3  13  17 2  19 2. = 2 2  5  17 2  19 3  23. 1 0 0 –1 1 0 –1 0 1. n 1 2 3 4 5 6 7 8 9. A n 1

By candice (97 views)

1 0 0 –1 1 0 –1 0 1

1 0 0 –1 1 0 –1 0 1

Totally Symmetric Self-Complementary Plane Partitions. 1 0 0 –1 1 0 –1 0 1. 1983. Totally Symmetric Self-Complementary Plane Partitions. 1 0 0 –1 1 0 –1 0 1. 1 0 0 –1 1 0 –1 0 1. Robbins’ Conjecture: The number of TSSCPP’s in a 2 n X 2n X 2 n box is. 1 0 0 –1

By lee-drake (77 views)

1 0 0 –1 1 0 –1 0 1

1 0 0 –1 1 0 –1 0 1

Percy A. MacMahon. Plane Partition. 1 0 0 –1 1 0 –1 0 1. Work begun in 1897. Plane partition of 75. 6 5 5 4 3 3. 1 0 0 –1 1 0 –1 0 1. # of pp’s of 75 = pp (75). Plane partition of 75. 6 5 5 4 3 3. 1 0 0 –1 1 0 –1 0 1.

By snez (0 views)

1 0 0 –1 1 0 –1 0 1

1 0 0 –1 1 0 –1 0 1

1979, Andrews counts cyclically symmetric plane partitions. 1 0 0 –1 1 0 –1 0 1. 1979, Andrews counts cyclically symmetric plane partitions. 1 0 0 –1 1 0 –1 0 1. 1979, Andrews counts cyclically symmetric plane partitions. 1 0 0 –1 1 0 –1 0 1.

By vlorenzo (0 views)

1 0 0 –1 1 0 –1 0 1

1 0 0 –1 1 0 –1 0 1

Percy A. MacMahon. Plane Partition. 1 0 0 –1 1 0 –1 0 1. Work begun in 1897. Plane partition of 75. 6 5 5 4 3 3. 1 0 0 –1 1 0 –1 0 1. # of pp’s of 75 = pp (75). Plane partition of 75. 6 5 5 4 3 3. 1 0 0 –1 1 0 –1 0 1.

By borka (75 views)

1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 8

1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 8

1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 8. 2 1 1 1 0 1 1 1 1 0 0 0 0 0 0 7. 3 0 1 1 1 1 1 1 1 1 0 0 0 0 0 8. 4 1 0 1 1 1 1 1 1 0 0 0 0 0 0 7. 5 0 0 1 1 1 0 1 0 0 0 0 0 0 0 4. 6 0 0 1 0 1 1 0 1 0 0 0 0 0 0 4. 7 0

By beckman (0 views)

EE1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0

EE1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0

E 3 143. E 4 1342. Graph Path Analytics (using pTrees). 1 0 0 0. Two-Level Stride=4, Edge pTrees. Two-Level Str=4, Unique Edge pTrees. 0 0 0 0. E 3 1. L1 3 14. L1 3 41. L1 3 31. L1 3 43. E 3 133. L2 3 4. E 3 142. L2 3 3. E 3 132. L2 3 2. E 3 13. E 3 121.

By hafford (0 views)

a 4 ‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0

a 4 ‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0

0 1. Vote histogram (so far). UF NNC Ptree Ex. 1 using 0-D Ptrees (sequences) a=a 5 a 6 a 1 ’a 2 ’a 3 ’a 4 ’=(000000).

By elaina (52 views)