[Story 49] Dasar Analitika Bisnis | [Semester 1] Part 4 : Quiz 1 - Dasar Analitika Bisnis | BAB : Logika dan Proposisi ; Logika dan Pembuktian

Subject : Dasar Analitika Bisnis

Tema : Dasar Analitika Bisnis | [Semester 1] Part 4 : Quiz 1 - Dasar Analitika Bisnis | BAB : Logika & Proposisi ; Logika dan Pembuktian

By : Mrs. Aruni Rahmaniar Purwanto, S. Si., M. Stat.


Questions :

1. Let p and q be the propositios.
p : you drive over 65 miles per hour
q : you get a spending ticket

Write these propositions using p and q and logical connectives (including negations).

a. You do not drive over 65 miles per hour

b. You drive over 65 miles per hour, but you do not get a spending ticket

c. You will get a spending ticket if you drive over 65 miles oer hour

d. If you do nog drive over 65 miles per hour, then you will not get a spending ticket

e. Driving over 65 miles per hour a sufficient for getting a spending ticket

f. You get a spending ticket, but you do not drive over 65 miles per hour

g. Wherever you get a spending ticket, you are driving over 65 miles per hour

2. Construct a truth table for each of these compound propositions.

a. (p ∨ q) 

b. (p ∧ q) ∧ r

c. (p ∧ q) ∨ r

d. p → (~q ∨ r)

e. ~p → (q → r)

f. (p → q) ∨ (~p → r)

(Kerjakan mandiri dulu, yaa, bestie. Jangan langsung nyontek kunci jawaban. Selamat menikmati) ‼️‼️ 🤗🤗🤗

Key answer : 

1.

a. You do not drive over 65 miles per hour

Ans. : (~p)

b. You drive over 65 miles per hour, but you do not get a spending ticket 

Ans. : (p → q)

c. You will get a spending ticket if you drive over 65 miles oer hour 

Ans. : (q → p)

d. If you do nog drive over 65 miles per hour, then you will not get a spending ticket

Ans. : (~p → ~q)

e. Driving over 65 miles per hour a sufficient for getting a spending ticket

Ans. : (p → q)

f. You get a spending ticket, but you do not drive over 65 miles per hour

Ans. : (q → ~p)

g. Wherever you get a spending ticket, you are driving over 65 miles per hour

Ans. : (p → q)

2.

a. (p ∨ q) 

b. (p ∧ q) ∧ r

c. (p ∧ q) ∨ r

d. p → (~q ∨ r)

e. ~p → (q → r)

f. (p → q) ∨ (~p → r)