Answer:
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Answer:
3/5
Step-by-step explanation:
Answer:
1) is not possible
2) P(A∪B) = 0.7
3) 1- P(A∪B) =0.3
4) a) C=A∩B' and P(C)= 0.3
b) P(D)= 0.4
Step-by-step explanation:
1) since the intersection of 2 events cannot be bigger than the smaller event then is not possible that P(A∩B)=0.5 since P(B)=0.4 . Thus the maximum possible value of P(A∩B) is 0.4
2) denoting A= getting Visa card , B= getting MasterCard the probability of getting one of the types of cards is given by
P(A∪B)= P(A)+P(B) - P(A∩B) = 0.6+0.4-0.3 = 0.7
P(A∪B) = 0.7
3) the probability that a student has neither type of card is 1- P(A∪B) = 1-0.7 = 0.3
4) the event C that the selected student has a visa card but not a MasterCard is given by C=A∩B' , where B' is the complement of B. Then
P(C)= P(A∩B') = P(A) - P(A∩B) = 0.6 - 0.3 = 0.3
the probability for the event D=a student has exactly one of the cards is
P(D)= P(A∩B') + P(A'∩B) = P(A∪B) - P(A∩B) = 0.7 - 0.3 = 0.4
The answer would be A. When using Cramer's Rule to solve a system of equations, if the determinant of the coefficient matrix equals zero and neither numerator determinant is zero, then the system has infinite solutions. It would be hard finding this answer when we use the Cramer's Rule so instead we use the Gauss Elimination. Considering the equations:
x + y = 3 and <span>2x + 2y = 6
Determinant of the equations are </span>
<span>| 1 1 | </span>
<span>| 2 2 | = 0
</span>
the numerator determinants would be
<span>| 3 1 | . .| 1 3 | </span>
<span>| 6 2 | = | 2 6 | = 0.
Executing Gauss Elimination, any two numbers, whose sum is 3, would satisfy the given system. F</span>or instance (3, 0), <span>(2, 1) and (4, -1). Therefore, it would have infinitely many solutions. </span>
Answer:
11=y
66=x
6=y
55=x
Step-by-step explanation:
Hope this helps
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