Answer: The given logical equivalence is proved below.
Step-by-step explanation: We are given to use truth tables to show the following logical equivalence :
P ⇔ Q ≡ (∼P ∨ Q)∧(∼Q ∨ P)
We know that
two compound propositions are said to be logically equivalent if they have same corresponding truth values in the truth table.
The truth table is as follows :
P Q ∼P ∼Q P⇔ Q ∼P ∨ Q ∼Q ∨ P (∼P ∨ Q)∧(∼Q ∨ P)
T T F F T T T T
T F F T F F T F
F T T F F T F F
F F T T T T T T
Since the corresponding truth vales for P ⇔ Q and (∼P ∨ Q)∧(∼Q ∨ P) are same, so the given propositions are logically equivalent.
Thus, P ⇔ Q ≡ (∼P ∨ Q)∧(∼Q ∨ P).
Answer:
1) -3/3
2) 1/2
3) 3/4
4) -2/1
5) 1/2
6) 4/1
Step-by-step explanation:
<u>RISE OVER RUN</u>
(Rise) Using the farthest dot on the left you count how much you rise till you align with the second dot <em>(depending on whether you go up or down determines + or -)</em>
(Run) Once aligned with the second dot count whilst 'running' towards it till you've reached it.<em> (If you go right it's +, if you go left it's -)</em>
Hard to explain, but it's really easy if you look up a tutorial.
Answer:
5/2
Step-by-step explanation:
5/2
Answer:
The correct option is;
(B) Yes, because sampling distributions of population proportions are modeled with a normal model.
Step-by-step explanation:
Here we have the condition for normality being that where we have a population with a given mean and standard deviation, while a sufficiently large sample is drawn from the population while being replaced, the distribution of the sample mean p will be distributed normally according to central limit theorem.
the answer is x is equal to -4
x=-4
