Help me please having trouble!?

Show that each of these conditional statements is a tautology by using truth tables. a) [¬p ∧ (p ∨ q)] → q b) [(p → q) ∧ (q → r)

nlexa [21]

Step-by-step explanation:

To show that a statement is a tautology using truth table - is to show that all the entries in the expression are truths T.

We can do this by taking each statement, expression by expression. For example, to show that

[~p ∧ (p ∨ q)] → q

is a tautology, knowing we have 3 columns, we have 2^3 = 8 rows. We start by putting putting truth values for p, q, and r respectively

Next, we find ~p, then find (p ∨ q), then find ~p ∧ (p ∨ q), before finally arriving at the required [~p ∧ (p ∨ q)] → q

TERMINOLOGIES AND SYMBOLS

- T means Truth

- F means False

- ~p means negation of p.

~p is F if p is T, and vice versa

- ∧ means conjunction.

p ∧ q is T only if p is T and q is T.

- ∨ means disjunction.

p ∨ q is T if either p or q is T.

- → is for conditional 'if then'

p → q is T if both p and q are T, or both p and q are F, or p is F and q is T, otherwise, it is F.

THE STEP BY STEP WORKINGS FOR THE STATEMENTS GIVEN ARE IN THE ATTACHMENT.

8 0

2 years ago

Read 2 more answers

Consider the system of linear equations.

Nadusha1986 [10]

Answer:

The system has one solution.

Both lines have the same y-intercept.

The solution is the intersection of the 2 lines.

Step-by-step explanation:

Lines are

$2y=x +10$

slope = 0.5

y intercept(put x =0) =5

$3y =3x +15$

slope = 1

y intercept(put x =0 ) =5

Both are linear equation thus they have only one solution. Also two non parallel lines meet only at one point.

If you solve these linear equation coordinate of point of intersection of line on graph will come.

4 0

3 years ago

I need help is it A B C or D

deff fn [24]

I would say c as the answer

8 0

2 years ago

Read 2 more answers

Discrete R.V. Assume a student shows up for EGR 280 completely unprepared and the instructor gives a pop quiz. Assume there are

Tom [10]

Answer:

a) p=0.2

b) probability of passing is 0.01696

.

c) The expected value of correct questions is 1.2

Step-by-step explanation:

a) Since each question has 5 options, all of them equally likely, and only one correct answer, then the probability of having a correct answer is 1/5 = 0.2.

b) Let X be the number of correct answers. We will model this situation by considering X as a binomial random variable with a success probability of p=0.2 and having n=6 samples. We have the following for k=0,1,2,3,4,5,6

$P(X=k) = \binom{n}{k}p^{k}(1-p)^{n-k} = \binom{6}{k}0.2^{k}(0.8)^{6-k}$ .

Recall that $\binom{n}{k}= \frac{n!}{k!(n-k)!}$ In this case, the student passes if X is at least four correct questions, then

$P(X\geq 4) = P(X=4)+P(X=5)+P(X=6)=\binom{6}{4}0.2^{4}(0.8)^{6-4}+\binom{6}{5}0.2^{5}(0.8)^{6-5}+\binom{6}{6}0.2^{6}(0.8)^{6-6}= 0.01696$

c)The expected value of a binomial random variable with parameters n and p is $E[X] = np$ . IN our case, n=6 and p =0.2. Then the expected value of correct answers is $6\cdot 0.2 = 1.2$