1.
The number of different values can be stored in x bits is: 2^x
The number of different values can be stored in 5 bits is: 2^5 = 32
one’s compliment format mean for the number of different values that can be stored changed to: 2^x – 1, because now 11111 and 00000 both represent 0.
highest number is 011111 (15), its magnitude is 15.
lowest number is 10000 (-15), its magnitude is 15.
two’s compliment format mean for the number of different values that can be stored changed to: 2^x , because now all binary represent different number.
highest number is 011111 (15), its magnitude is 15.
lowest number is 10000 (-16), its magnitude is 16.
binary
integer
one
two
00000
0
0
0
00001
1
1
1
00010
2
2
2
00011
3
3
3
00100
4
4
4
00101
5
5
5
00110
6
6
6
00111
7
7
7
01000
8
8
8
01001
9
9
9
01010
10
10
10
01011
11
11
11
01100
12
12
12
01101
13
13
13
01110
14
14
14
01111
15
15
15
10000
16
-15
-16
10001
17
-14
-15
10010
18
-13
-14
10011
19
-12
-13
10100
20
-11
-12
10101
21
-10
-11
10110
22
-9
-10
10111
23
-8
-9
11000
24
-7
-8
11001
25
-6
-7
11010
26
-5
-6
11011
27
-4
-5
11100
28
-3
-4
11101
29
-2
-3
11110
30
-1
-2
11111
31
-0
-1
2.
Bits per character is 100 * 8 / 100 = 8 (bits).
Number of possible symbols could be represented in this way: 2^8 = 256.
There are 26 lowercase letters, and because 2^4 < 26 < 2^5, so they need at least 5 bits.
Upper and lower together is 52, because 2^5 < 26 < 2^6, so they need at least 6 bits.
There are 52 letters, plus 10 numbers and 33 punctuations, in total it is 95, because 2^6 < 95 < 2^7, so they need at least 7 bits.
4.
5 + 5 = 101 + 101 = 1010 = 10
8 + 7 = 1000 + 111 = 1111 = 15
9 - 3 = 1001 - 11 = 110 = 6
7 - 5 = 111 - 101 = 10 = 2
5.
1011 = 23 + 21 + 20 = 8 + 2 + 1 = 11
1 0 1 0 0 0 0 = 26 + 24 = 64 + 16 = 80
1 1 00 1 1 1 1 = 27 + 26 + 23 + 22 + 21 + 20
= 128 + 64 + 8 + 4 + 2 + 1 = 207
0 1 1 1 1 0 0 0 = 26 + 25 + 24 + 23 = 64 + 32 + 16 + 8 = 120
1 0 1 0 0 00 0 1 1 1 0 1 000 = 215 + 213 + 27 + 26 + 25 + 23
= 32768 + 8192 + 128 + 64 + 32 + 8 = 41192
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