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Ainat [17]
2 years ago
15

HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HEL

P HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP

Computers and Technology
2 answers:
soldi70 [24.7K]2 years ago
6 0

Answer:

B

Explanation:

AURORKA [14]2 years ago
5 0

Answer:

I should be C! :D

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Read 2 more answers
let m be a positive integer with n bit binary representation an-1 an-2 ... a1a0 with an-1=1 what are the smallest and largest va
Ahat [919]

Answer:

Explanation:

From the given information:

a_{n-1} , a_{n-2}...a_o in binary is:

a_{n-1}\times 2^{n-1}  + a_{n-2}}\times 2^{n-2}+ ...+a_o

So, the largest number posses all a_{n-1} , a_{n-2}...a_o  nonzero, however, the smallest number has a_{n-2} , a_{n-3}...a_o all zero.

∴

The largest = 11111. . .1 in n times and the smallest = 1000. . .0 in n -1 times

i.e.

(11111111...1)_2 = ( 1 \times 2^{n-1} + 1\times 2^{n-2} + ... + 1 )_{10}

= \dfrac{1(2^n-1)}{2-1}

\mathbf{=2^n -1}

(1000...0)_2 = (1 \times 2^{n-1} + 0 \times 2^{n-2} + 0 \times 2^{n-3} + ... + 0)_{10}

\mathbf {= 2 ^{n-1}}

Hence, the smallest value is \mathbf{2^{n-1}} and the largest value is \mathbf{2^{n}-1}

3 0
2 years ago
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