Then your amount was reduced to $15 less
Answer: 0.05
Step-by-step explanation:
Let M = Event of getting an A in Marketing class.
S = Event of getting an A in Spanish class,
i.e. P(M) = 0.80 , P(S) = 0.60 and P(M∩S)=0.45
Required probability = P(neither M nor S)
= P(M'∩S')
= P(M∪S)' [∵P(A'∩B')=P(A∪B)']
=1- P(M∪S) [∵P(A')=1-P(A)]
= 1- (P(M)+P(S)- P(M∩S)) [∵P(A∪B)=P(A)+P(B)-P(A∩B)]
= 1- (0.80+0.60-0.45)
= 1- 0.95
= 0.05
hence, the probability that Helen does not get an A in either class= 0.05
Answer:
-22
Step-by-step explanation:
6j-5k=11
5j-6k=22
-----------------
5(6j-5k)=5(11)
-6(5j-6k)=-6(22)
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30j-25k=55
-30j+36k=-132
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11k=-77
k=-77/11
k=-7
6j-5(-7)=11
6j+35=11
6j=11-35
6j=-24
j=-24/6
j=-4
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2j+2k=2(-4)+2(-7)=-8-14=-22
Answer:
The probability is found:
P = 24/3024 = 1/126
Step-by-step explanation:
To find total number of outcomes, we have to find permutation of 9 things taken 4 at a time:
P(9,4) = 9! / (9-3)!
P(9,4) =362880/120
P(9,4) = 3024
Number of even numbers from 0 to 9 = 4
To find desirable number of outcome, find permutation of 4 things taken 4 at a time.
P(4,4) = 4! / (4-4)!
P(4,4) = 24/1
P(4,4) = 24
The probability that the lock consists of all even digits is
P = No. of desirable outcomes / No. of total outcomes
P = 24/3024
Answer:
Step-by-step explanation:
