1.20 + 0.34 = 1.54
That is the answer.
Answer:
it should be the negative repirocal of -2 so 1/2
Answer:
a. 0.76
b. 0.23
c. 0.5
d. p(B/A) is the probability that given that a student has a visa card, they also have a master card
p(A/B) is the probability that given a student has a master card, they also have a visa card
e. 0.35
f. 0.31
Step-by-step explanation:
a. p(AUBUC)= P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AnBnC)
=0.6+0.4+0.2-0.3-0.11-0.1+0.07= 0.76
b. P(AnBnC')= P(AnB)-P(AnBnC)
=0.3-0.07= 0.23
c. P(B/A)= P(AnB)/P(A)
=0.3/O.6= 0.5
e. P((AnB)/C))= P((AnB)nC)/P(C)
=P(AnBnC)/P(C)
=0.07/0.2= 0.35
f. P((AUB)/C)= P((AUB)nC)/P(C)
=(P(AnC) U P(BnC))/P(C)
=(0.11+0.1)/0.2
=0.21/0.2 = 0.31
As follows
r-6=7
+6 +6
r= 13
I wrote the domain and range in interval notation, not sure if that’s how you’re asked to do it. This is how I was taught to state the domain/range of a graph.
