decimal octal binary logic gates Quiz Practice…

11 — 1*2^3 + 0*2^2 + 1*2^1 + 1*2^0 = 8 + 0 + 2 + 1 = 11

1000 — 8 = 1*2^3 + 0*2^2 + 0*2^1 + 0*2^0 = 1000

AND gate — An AND gate requires all inputs to be high for a high output.

10011 — 2 in octal is 010 in binary, and 3 in octal is 011 in binary; hence 010 011 -> 10011

NAND gate — A NAND gate is the inverted output of an AND gate.

Question 1 of 5

What is the decimal equivalent of the binary number 1011?

1*2^3 + 0*2^2 + 1*2^1 + 1*2^0 = 8 + 0 + 2 + 1 = 11

Question 2 of 5

8 = 1*2^3 + 0*2^2 + 0*2^1 + 0*2^0 = 1000

Question 3 of 5

An AND gate requires all inputs to be high for a high output.

Question 4 of 5

2 in octal is 010 in binary, and 3 in octal is 011 in binary; hence 010 011 -> 10011

Question 5 of 5

A NAND gate is the inverted output of an AND gate.