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Questions by Nishant Arora

1 votes
2 answers
1
MadeEasy Test Series: Linear Algebra - Eigen Value
can we use some identity to find value of ab+by+ya???without finding its eigen values and then multiplying..earlier i have read it somewhere but i forget it.
can we use some identity to find value of ab+by+ya???without finding its eigen values and then multiplying..earlier i have read it somewhere but i forget it.
0 votes
0 answers
3
ME FLT-1 Q-10
i have used hypergeometric distribution but MADE Easy have used binomial theorem..we should use binomial when replacement are done na? my ans.0.6 made easy ans:0.81
i have used hypergeometric distribution but MADE Easy have used binomial theorem..we should use binomial when replacement are done na?my ans.0.6made easy ans:0.81
3 votes
1 answer
4
made easy technical
i think option should be B if it is left skewed..same type questionhttps://gateoverflow.in/974/gate2006-13
i think option should be B if it is left skewed..same type questionhttps://gateoverflow.in/974/gate2006-13
0 votes
0 answers
6
memory management
A computer system has a 36-bit virtual address space with a page size of 8K, and 4 bytes per page table entry.What is the maximum size of addressable physical memory in this system?
A computer system has a 36-bit virtual address space with a page size of 8K, and 4 bytes per page table entry.What is the maximum size of addressable physical memory in t...
1 votes
1 answer
7
http://gateoverflow.in/31625/gate1990-what-is-the-asymptotic-behaviour-of-t-n
while solving recurrence i got this series..now how to proceed further? $1+\frac{n+1}{2}+\frac{(n+1)n}{2^{2}}+\frac{(n+1)n(n-1))}{2^{3}}+..\frac{(n+1)n(n-1)(n-2)(n-3)...3.2.1}{2^{n+1}}$
while solving recurrence i got this series..now how to proceed further?$1+\frac{n+1}{2}+\frac{(n+1)n}{2^{2}}+\frac{(n+1)n(n-1))}{2^{3}}+..\frac{(n+1)n(n-1)(n-2)(n-3)...3....