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1992-06-19
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mode3
O11
text/13 5 0 1 1
For this course, you don't have
to learn binary counting - just
know that this is how the com-
puter uses combinations of on and
off to count.
However, if you want to give
it a try, name this number:
~
O9
R50 80 243 87
L150 81 150 87
L101 79 101 79
L102 81 102 86
L198 79 198 79
L198 81 198 86
L219 79 219 79
L220 80 220 86
L173 79 173 79
L173 80 173 86
L126 79 126 79
L126 81 126 86
L75 80 75 80
L75 80 75 86
O11
text/5 96 0 1 1
Here's a hint: Remember that the
rightmost position, if "on" is
worth 1. The next position to the
left is worth twice that, if "on"
is therefore worth 2, and the next
one to the left of that is worth 4
and then the next is 8, and so on.
Just add the ones that are "on"
together.
~
KC 48 79 246 89 387 15
O10
text 568 31 0 1 1
1
~
text 546 31 0 1 1
2
~
text 520 31 0 1 1
4
~
text 498 31 0 1 1
8
~
text 470 31 0 1 1
16
~
text 418 31 0 1 1
64
~
text 385 40 1 1 0
128
~
text 442 40 1 1 0
32
~
O12
L397 27 397 37
L425 25 425 25
L425 25 425 28
L451 25 451 25
L451 25 451 38
L477 26 477 26
L477 25 477 29
L499 25 499 25
L499 25 499 28
L522 25 522 25
L523 25 523 28
L547 24 547 24
L548 25 548 28
L569 25 569 25
L569 25 569 28
O11
text 341 50 1 1 0
The bits have these values if
turned on. So to get the decimal
value of the byte, add up the
values of the "on" bits.
~
O10
text 462 96 0 1 1
128
~
text 478 105 0 1 1
+16
~
text 478 114 0 1 1
~
text 485 114 0 1 1
+8
~
text 485 123 0 1 1
+1
~
O11
L463 133 509 133
O10
text 469 134 0 1 1
= ?
~
O11
text 400 146 0 1 1
Answer on next page.
~
mode3
~
KY 424 5 569 15
KY 80 76 222 79
O0
P547 24
P522 25
O13
R51 81 73 86
R127 81 148 86
R152 81 172 86
R221 81 241 86
F67 83 1
F137 83 1
F167 83 1
F235 84 1
KC 48 79 246 89 387 15
mode3