Answer:
a. 0.9
b. 0.024
c. 0.06
d. 0.2
e. 0.336
Step-by-step explanation:
Since A, B and C are mutually exclusive then P (A U B U C)=P(A)+P(B)+P(C) and P(A∩B∩C)=P(A)*P(B)*P(C).
a.
P (A U B U C)=P(A)+P(B)+P(C)
P (A U B U C)=0.2+0.3+0.4=0.9
b.
P(A∩B∩C)=P(A)*P(B)*P(C)
P(A∩B∩C)=0.2*0.3*0.4=0.024
c.
P(A∩B)=P(A)*P(B)
P(A∩B)=0.2*0.3=0.06
d.
P[(AUB)∩C]=P(AUB)*P(C)
P(AUB)=P(A)+P(B)=0.2+0.3=0.5
P[(AUB)∩C]=0.5*0.4=0.20
e.
P(A')=1-P(A)=1-0.2=0.8
P(B')=1-P(B)=1-0.3=0.7
P(C')=1-P(C)=1-0.4=0.6
P(A'∩B'∩ C')=P(A')*P(B')*P(C')
P(A'∩B'∩ C')=0.8*0.7*0.6=0.336
I think it’s 8 because when i divide 2.93 by 1/4 it’s 11.3948393 and it says how many can he make so it has to b 8
Find the first few
coefficients in the power series 4/1+4x^2.
Then we have:
4/ (1 – (-4x^2) = 4 Σ^ (n
≥ 0) (-4x ^2) ^n
= 4 (1 – 4x^2 + 4x^2 x^4 - … + (-4x ^2) ^n) These are the
first few coefficient of the series.
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