Commutative but not associative
Both commutative and associative
Associative but not commutative
Neither commutative nor associative
8 / (2e3)
9 / (2e3)
17 / (2e3)
26 / (2e3)
PR(X < 3) = Pr(x = 0) + Pr(x = 1) + Pr(x = 2) = f(0, 3) + f(1, 3) + f(2, 3) Put [Tex]\lambda[/Tex] = 3 and k = 0, 1, 2 in the formula given at http://en.wikipedia.org/wiki/Poisson_distribution#Definition = 17 / (2e3)
First of all, you should know the basic properties of determinants before approaching For these kind of problems. 1) Applying any row or column transformation does not change the determinant 2) If you interchange any two rows, sign of the determinant will change
A = | 1 x x^2 | | 1 y y^2 | | 1 z z^2 |
To prove option (b)
=> Apply column transformation C2 -> C2+C1
C3 -> C3+C1
=> det(A) = | 1 x+1 x^2+1 | | 1 y+1 y^2+1 | | 1 z+1 z^2+1 |
To prove option (c),
=> Apply row transformations R1 -> R1-R2
R2 -> R2-R3
=> det(A) = | 0 x-y x^2-y^2 | | 0 y-z y^2-z^2 | | 1 z z^2 |
To prove option (d),
=> Apply row transformations R1 -> R1+R2
R2 -> R2+R3
=> det(A) = | 2 x+y x^2+y^2 | | 2 y+z y^2+z^2 | | 1 z z^2 |
This solution is contributed by Anil Saikrishna Devarasetty .
To insert 50, we will have to traverse all nodes. 10 \ 20 \ 30 \ 40
S->aB B->bC C->cdThe Right Most Derivation for the above is:
S -> aB ( Reduction 3 ) -> abC ( Reduction 2 ) -> abcd ( Reduction 1 )We can see here that no production is for unit or epsilon. Hence 3 reductions here. We can get less number of reductions with some other grammar which also does't produce unit or epsilon productions,
S->abA A-> cdThe Right Most Derivation for the above as:
S -> abA ( Reduction 2 ) -> abcd ( Reduction 1 )Hence 2 reductions. But we are interested in knowing the maximum number of reductions which comes from the 1st grammar. Hence total 3 reductions as maximum, which is ( n - 1) as n = 4 here. Thus, Option B.
This algorithm is equivalent to the first-come-first-serve algorithm
This algorithm is equivalent to the round-robin algorithm.
This algorithm is equivalent to the shortest-job-first algorithm..
This algorithm is equivalent to the shortest-remaining-time-first algorithm