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Rudik [331]
3 years ago
5

A _____ object is the instance of a class that receives a request from another object. Select one: a. client b. server c. contra

ct d. provider e. CRC
Computers and Technology
1 answer:
creativ13 [48]3 years ago
5 0

Answer:

b. server

Explanation:

A server serves responses to requests from client objects.

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To employ an access key, press and hold down the ____ key as you tap the access key.
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To employ an access key, press and hold down the Alt key as you tap the access key.
4 0
2 years ago
Construct an algorithm to print the first 20 numbers in the Fibonacci series (in mathematics) thanks
ss7ja [257]

Answer:

0+1=1

1+1=2

1+2=3

2+3=5

3+5=8

5+8=13

Explanation:

// C++ program to print

// first n Fibonacci numbers

#include <bits/stdc++.h>

using namespace std;

 

// Function to print

// first n Fibonacci Numbers

void printFibonacciNumbers(int n)

{

   int f1 = 0, f2 = 1, i;

 

   if (n < 1)

       return;

   cout << f1 << " ";

   for (i = 1; i < n; i++) {

       cout << f2 << " ";

       int next = f1 + f2;

       f1 = f2;

       f2 = next;

   }

}

 

// Driver Code

int main()

{

   printFibonacciNumbers(7);

   return 0;

}

 

8 0
3 years ago
Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter:
horsena [70]

Answer:

Given Data:

Rotation Speed = 7200 rpm

No. of tracks on one side of platter = 30000

No. of sectors per track = 600

Seek time for every 100 track traversed = 1 ms

To find:

Average Seek Time.

Average Rotational Latency.

Transfer time for a sector.

Total Average time to satisfy a request.

Explanation:

a) As given, the disk head starts at track 0. At this point the seek time is 0.

Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

29999 / 2 = 14999.5 ms

As the seek time is one ms for every hundred tracks traversed.  So the seek time for 29,999 tracks traversed is

14999.5 / 100 = 149.995 ms

b) The rotations per minute are 7200

1 min = 60 sec

7200 / 60 = 120 rotations / sec

Rotational delay is the inverses of this. So

1 / 120 = 0.00833 sec

          = 0.00833 * 100

          = 0.833 ms

So there is  1 rotation is at every 0.833 ms

Average Rotational latency is one half the amount of time taken by disk to make one revolution or complete 1 rotation.

So average rotational latency is: 1 / 2r

8.333 / 2 = 4.165 ms

c) No. of sectors per track = 600

Time for one disk rotation = 0.833 ms

So transfer time for a sector is: one disk revolution time / number of sectors

8.333 / 600 = 0.01388 ms = 13.88 μs

d)  Total average time to satisfy a request is calculated as :

Average seek time + Average rotational latency + Transfer time for a sector

= 149.99 ms + 4.165 ms + 0.01388 ms

= 154.168 ms

4 0
3 years ago
A + B (AC (B +C’) D) = A + BD
zmey [24]
Yo what is that i’m confused
5 0
3 years ago
Read 2 more answers
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