Question 1202032: Let p be Ricky love Lucy and let q be Lucy loves Ricky. Give a verbal translation.
a Pv~Q
b ~P^~Q
c~(P^Q)
2. Construct truth table for the following
a Pv~Q
b Pv(Q^r)
c ~P>Q<>PvQ
3.in survey of 100 students,the members various courses were found to be English 60, mathematics 40, chemistry 50 English and mathematics 30, English and chemistry 35, mathematics and chemistry 35 and courses in all three areas 25
a.how many students were taking mathematics, but neither English nor chemistry
b.how many were taking mathematics and chemistry but not English.
c.how many were taking English and chemistry but not mathematics
Answer by ikleyn(52786) (Show Source):
You can put this solution on YOUR website! .
in survey of 100 students, the members various courses were found to be English 60, mathematics 40, chemistry 50
English and mathematics 30, English and chemistry 35, mathematics and chemistry 35
and courses in all three areas 25
a. how many students were taking mathematics, but neither English nor chemistry
b. how many were taking mathematics and chemistry but not English.
c. how many were taking English and chemistry but not mathematics
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In this post, I will solve the third problem ONLY from your post.
In this problem, you are given a universe set of all 100 students
and its three basic subsets
E of 60 students learning English
M of 40 students learning Math
C of 50 students learning Chemistry.
You are also given in-pair intersections of these three subsets
EM of 30 students learning English and Math
EC of 35 students learning English and Chemistry
MC of 35 students learning Math and Chemistry
You are also given the triple intersection EMC of 25 students learning all three subjects.
(a) how many students were taking mathematics, but neither English nor chemistry ?
To answer this question, we subtract the quantities EM and MC from M, first.
But doing it, we subtract the amount in the triple intersection EMC twice.
Therefore, to compensate it, we add EMC at the end, so out formula is
n(M_only) = n(M) - n(EM) - n(MC) + n(EMC) = 40 - 30 - 35 + 25 = 0. ANSWER
(b) how many were taking mathematics and chemistry but not English ?
To answer this question, we take the quantity n(MC) and subtract quantity n(EMC) from it
n(MC_only) = n(MC) - n(MCE) = 35 - 25 = 10. ANSWER
(c) how many were taking English and chemistry but not mathematics ?
This question is very similar to (b), and the answer to it is very similar to (b), too.
To answer this question, we take the quantity n(EC) and subtract quantity n(EMC) from it
n(EC_only) = n(EC) - n(MCE) = 35 - 25 = 10. ANSWER
Solved. // I answered all your questions to this problem.
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In the future, NEVER pack more than one problem per post
(for the sake of your benefits . . . )
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