# Recent posts tagged go-classroom

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Probability that buckets other than $1$ are selected $= \dfrac{n-1}{n}$ This should happen $k$ times and each of the $k$ events are independent so $\dfrac{(n-1)^k}{n^k}$ When $k=1,$ probability of no collision $= \dfrac{n}{n}$ (for only one insert, there can ... $k>n+1$ probability of first collision $= 0$ (as it should have definitely happened in one of the previous $(n+1)$ insertions.
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if you solve this you will get XY' + Y'Z + XY (this can be simplified to X + Y'Z) with min terms as (1,4,5,6,7) and option A has the same min terms so option A is equivalent to XY' + Y'Z + XY Ans (A)
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option A Let P(x)=All are multiple of 4 Q(X)= All are even numbers if a &isin; P(X) then a must &isin; Q(X)
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The number of surjective functions defined from $A$ to $B$ where |A| = 5, |B| = 4, is _______
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$\text{Size of each segment} = \frac{2^{16}}{8}=2^{13}$ Let the size of page be $2^k$ bytes We need a page table entry for each page. For a segment of size $2^{13}$, number of pages required will be $2^{13-k}$ and so we need $2^{13-k}$ page table entries. Now ... $\text{Size of each page table entry} = 2$ bytes $=16$ bits $\text{Number of bits left for aging} = 16-12 = 4$ bits
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Ans: because we have given wavy edges form MST So, for verification of option A we have to check that with MST how many cost to reach at a->b then we will get a->e->d->b = -2+5+3 = 6 so in given option a with cost(a,b)>= 6 this is posible coz , cost ... making MST so cost must be >6 not equal to 6 so option A is Need Not HOLD. Like wise if you check for other option then enequality is holding...
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see my comment on the question. That's a mistake actually. Unit productions are of type A-->B. there shouldn't be any terminal on the RHS.
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In the diagram shown below, $L1$ is an Ethernet LAN and $L2$ is a Token-Ring LAN. An $IP$ packet originates from sender $S$ and traverses to $R$, as shown. The links within each $\text{ISP}$ and across the two $\text{ISP}$s, are all point-to-point optical links ... of the $\text{TTL}$ field is $32$. The maximum possible value of the $\text{TTL}$ field when $R$ receives the datagram is _______.
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