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Karo-lina-s [1.5K]
2 years ago
12

Samantha was calculating a mathematical formula on an electronic spreadsheet. She used multiple values to recalculate the formul

a. Where will the CPU store data for easy access and quick retrieval during these computations?
A) read-only memory 
B) random- access memory 
C)  Hard disk 
D) Compact disk 
Computers and Technology
2 answers:
EastWind [94]2 years ago
8 0

Answer: B) random- access memory(RAM)

Explanation: Random-access memory is the memory unit  of the operating system that operates at high speed and can be accessed randomly The data stored in this memory can be revived till the system is on position.

According to the case mentioned in the question,Samantha can retrieve the formula stored in the form of data from RAM and use it whenever she requires it for calculation till the computer is on.This shows the volatile nature of RAM.

Other options are incorrect because read-only memory(ROM) can store data which can be only read, hard disk is non-volatile disk that stores data and compact disk(CD) stores data using light .Thus, the correct option is option(B).

MrMuchimi2 years ago
4 0
B. random-access memory
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emmainna [20.7K]

Answer:

They can be used to compare a certain region before and after disaster.

Explanation:

4 0
2 years ago
Suppose a host has a 1-MB file that is to be sent to another host. The file takes 1 second of CPU time to compress 50%, or 2 sec
kozerog [31]

Answer: bandwidth = 0.10 MB/s

Explanation:

Given

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.50 MB / Bandwidth)

Transfer Size = 0.50 MB

Total Time = Compression Time + RTT + (0.50 MB /Bandwidth)

Total Time = 1 s + RTT + (0.50 MB / Bandwidth)

Compression Time = 1 sec

Situation B:

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.40 MB / Bandwidth)

Transfer Size = 0.40 MB

Total Time = Compression Time + RTT + (0.40 MB /Bandwidth)

Total Time = 2 s + RTT + (0.40 MB / Bandwidth)

Compression Time = 2 sec

Setting the total times equal:

1 s + RTT + (0.50 MB / Bandwidth) = 2 s + RTT + (0.40 MB /Bandwidth)

As the equation is simplified, the RTT term drops out(which will be discussed later):

1 s + (0.50 MB / Bandwidth) = 2 s + (0.40 MB /Bandwidth)

Like terms are collected:

(0.50 MB / Bandwidth) - (0.40 MB / Bandwidth) = 2 s - 1s

0.10 MB / Bandwidth = 1 s

Algebra is applied:

0.10 MB / 1 s = Bandwidth

Simplify:

0.10 MB/s = Bandwidth

The bandwidth, at which the two total times are equivalent, is 0.10 MB/s, or 800 kbps.

(2) . Assume the RTT for the network connection is 200 ms.

For situtation 1:  

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 1 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.50 MB

Total Time = 1.2 sec + 5 sec

Total Time = 6.2 sec

For situation 2:

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 2 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.40 MB

Total Time = 2.2 sec + 4 sec

Total Time = 6.2 sec

Thus, latency is not a factor.

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1 year ago
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2 years ago
Write a copy constructor for carcounter that assigns origcarcounter.carcount to the constructed object's carcount. sample output
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#include <iostream>

using namespace std;

class CarCounter {

  public:

     CarCounter();

     CarCounter(const CarCounter& origCarCounter);

     void SetCarCount(const int count) {

         carCount = count;

     }

     int GetCarCount() const {

         return carCount;

     }

  private:

     int carCount;

};

CarCounter::CarCounter() {

  carCount = 0;

  return;

}

CarCounter::CarCounter(const CarCounter &p){

carCount = p.carCount;

}

void CountPrinter(CarCounter carCntr) {

  cout << "Cars counted: " << carCntr.GetCarCount();

  return;

}

int main() {

  CarCounter parkingLot;

  parkingLot.SetCarCount(5);

  CountPrinter(parkingLot);

  return 0;

}

Sample output:  

Cars Counted: 5

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