A
B
C
D

# Q2: One of DeMorgan's theorems states that . Simply stated, this means that logically there is no difference between:

A a NAND gate and an AND gate with a bubbled output
B a NOR gate and an AND gate with a bubbled output
C a NOR gate and a NAND gate with a bubbled output
D a NAND gate and an OR gate with a bubbled output

# Q3: Which of the examples below expresses the distributive law of Boolean algebra?

A A • (B • C) = (A • B) + C
B A + (B + C) = (A • B) + (A • C)
C A • (B + C) = (A • B) + (A • C)
D (A + B) + C = A + (B + C)

# Q4: The observation that a bubbled input OR gate is interchangeable with a bubbled output AND gate is referred to as:

A a Karnaugh map
B DeMorgan's second theorem
C the commutative law of addition
D the associative law of multiplication

A NAND
B NOR
C AND
D OR

# Q6: Which of the following expressions is in the sum-of-products (SOP) form?

A Y = (A + B)(C + D)
B Y = AB(CD)
C
D

# Q7: The systematic reduction of logic circuits is accomplished by:

A symbolic reduction
B TTL logic
C using Boolean algebra
D using a truth table

# Q8: Logically, the output of a NOR gate would have the same Boolean expression as a(n):

A NAND gate immediately followed by an INVERTER
B OR gate immediately followed by an INVERTER
C AND gate immediately followed by an INVERTER
D NOR gate immediately followed by an INVERTER

# Q9: Which of the examples below expresses the commutative law of multiplication?

A A + B = B + A
B A • B = B + A
C A • (B • C) = (A • B) • C
D A • B = B • A

# Q10: Which statement below best describes a Karnaugh map?

A It is simply a rearranged truth table.
B The Karnaugh map eliminates the need for using NAND and NOR gates.
C Variable complements can be eliminated by using Karnaugh maps.
D A Karnaugh map can be used to replace Boolean rules.

# Q11: The commutative law of addition and multiplication indicates that:

A the way we OR or AND two variables is unimportant because the result is the same
B we can group variables in an AND or in an OR any way we want
C an expression can be expanded by multiplying term by term just the same as in ordinary algebra
D the factoring of Boolean expressions requires the multiplication of product terms that contain like variables

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