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Is not p and q equivalent to not p and not q?
Think of 'not' as being an inverse. Not 1 = 0. Not 0 = 1. Using boolean algebra we can look at your question. 'and' is a test. It wants to know if BOTH P and Q are the same and if they are 1 (true). If they are not the same, or they are both 0, then the result is false or 0. not P and Q is rewritten like so: (P and Q)' = X not P and not Q is rewritten like: P' and Q' = X (the apostrophe is used for not) We will construct a truth table for each and compare the output. If the output is the same, then you have found your equivalency. Otherwise, they are not equivalent. P and Q are the inputs and X is the output. P Q | X P Q | X ------ 0 0 | 1 0 0 | 1 0 1 | 1 0 1 | 0 1 0 | 1 1 0 | 0 1 1 | 0 1 1 | 0 Since the truth tables are not equal, not P and Q is not equivalent to not P and not Q. Perhaps you meant "Is NOT(P AND Q) equivalent to NOT(P) AND NOT(Q)?" NOT(P AND Q) can be thought of intuitively as "Not both P and Q." Which if you think about, you can see that it would be true if something were P but not Q, Q but not P, and neither P nor Q-- so long as they're not both true at the same time. Now, "NOT(P) AND NOT(Q)" is clearly _only_ true when BOTH P and Q are false. So there are cases where NOT(P AND Q) is true but NOT(P) AND NOT(Q) is false (an example would be True(P) and False(Q)). NOT(P AND Q) does have an equivalence however, according to De Morgan's Law. The NOT can be distributed, but in doing so we have to change the "AND" to an "OR". NOT(P AND Q) is equivalent to NOT(P) OR NOT(Q)
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If B is between P and Q?
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