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juin [17]
3 years ago
14

Suppose a file system can have three disk allocation strategies, contiguous, linked, and indexed. We have just read the informat

ion for a file from its parent directory. For contiguous and linked allocation, this gives the address of the first block, and for indexed allocation this gives the address of the index block. Now we want to read the 10th data block into the memory. How many disk blocks (N) do we have to read for each of the allocation strategies?
Computers and Technology
1 answer:
otez555 [7]3 years ago
6 0

Answer:

Since we are using three disk allocation strategies,continuous linked and indexed,therof to process and read allocation strategies we need 99830 disk blocks.

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Answer:

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Explanation:

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The replacer parameter of the stringify method accepts a/an ______________ or an array.
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What facilitates the automation and management of business processes and controls the movement of work through the business proc
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Answer:

D. Workflow management systems

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3 years ago
You are given a string of n characters s[1 : : : n], which you believe to be a corrupted text document in which all punctuation
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Answer: provided in the explanation section

Explanation:

Given that:

Assume D(k) =║ true it is [1 : : : k] is valid sequence  words or false otherwise

  • To determine D(n)

now the sub problem s[1 : : : k] is a valid sequence of words IFF s[1 : : : 1] is a valid sequence of words and s[ 1 + 1 : : : k] is valid word.

So, from here we have that D(k) is given by the following recorance relation:

D(k) = ║ false maximum (d[l]∧DICT(s[1 + 1 : : : k]) otherwise

Algorithm:

Valid sentence (s,k)

D [1 : : : k]             ∦ array of boolean variable.

for a ← 1 to m

do ;

d(0) ← false

for b ← 0 to a - j

for b ← 0 to a - j

do;

if D[b] ∧ DICT s([b + 1 : : : a])

d (a) ← True

(b). Algorithm Output

      if D[k] = = True

stack = temp stack           ∦stack is used to print the strings in order

c = k

while C > 0

stack push (s [w(c)] : : : C] // w(p) is the position in s[1 : : : k] of the valid world at // position c

P = W (p) - 1

output stack

= 0 =

cheers i hope this helps !!!

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