Question

Which of the following binary (base-2) numbers is LARGEST?

Answer 1

Answer:

A. 11000000

Explanation:

A. 11000000 is 192

B. 01111111 is 127

C. 00000001 is 1

D. 10111111 is 191

Therefore, making A the largest.

Answer 2

The base 2 value which is largest from the options given is 11000000

Converting the values to base 10 :

1. 11000000

• $(1 \times {2}^{7}) + (1 \times {2}^{6}) + (0 \times {2}^{5}) + (0 \times {2}^{4}) + (0 \times {2}^{3}) + (0 \times {2}^{2}) + (0 \times {2}^{1}) + (0 \times {2}^{0} ) = 128 + 64 = 192$

2. 01111111

• $(0 \times {2}^{7}) + (1 \times {2}^{6}) + (1 \times {2}^{5}) + (1 \times {2}^{4}) + (1 \times {2}^{3}) + (1 \times {2}^{2}) + (1 \times {2}^{1}) + (1 \times {2}^{0} ) = 0 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 127$

3. 00000001

• $(0 \times {2}^{7}) + (0 \times {2}^{6}) + (0 \times {2}^{5}) + (0 \times {2}^{4}) + (0 \times {2}^{3}) + (0 \times {2}^{2}) + (0 \times {2}^{1}) + (1 \times {2}^{0} ) = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 1 = 1$

4. 10111111

• $(1 \times {2}^{7}) + (0 \times {2}^{6}) + (1 \times {2}^{5}) + (1 \times {2}^{4}) + (1 \times {2}^{3}) + (1 \times {2}^{2}) + (1 \times {2}^{1}) + (1 \times {2}^{0} ) = 128 + 0 + 32 + 16 + 8 + 4 + 2 + 1 = 191$

• The largest base 2 value obtained by comparing the equivalent base 10 values is 11000000

Learn more:https://brainly.com/question/23939074