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2 years ago

What explains the discrepancy between the number of bytes you can

Computers and Technology

1 answer:

sergejj [24]2 years ago

8 0

Answer:

This is because, the advertisers report the storage space in decimal while the computer reads the storage space in binary.

Explanation:

The advertisers report the storage space in decimal or base 10 because humans count in decimal, whereas the computer reports the storage space in binary or base 2.

Since the computer storage is in bytes and 8 bits equal 1 byte, It is easier to write a storage space of 1 kB as 1000 B but it is actually supposed to be 1024 B(2¹⁰). So, there is a discrepancy of 1024 B - 1000 B = 24 B.

As we go higher, the discrepancy increases. For example, 1 MB is advertised as 1000 kB = 1000000 B but is actually supposed to be 1024 kB = 1024 × 1024 B = 1048576 B. So, there is a discrepancy of 1048576 B - 1000000 B = 48576 B.

So, the actual number of bytes on the storage device is actually less than that reported due to the different number systems in which they are reported in. This discrepancy is less in memory cards or flash drives though in which the stated value of storage capacity might be the actual storage size.

Note that the base 10 or decimal system was chosen by advertiser since this is what consumers understand.

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2 years ago

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3 years ago

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2 years ago

given an array of integers a, your task is to count the number of pairs i and j (where 0 ≤ i < j < a.length), such that a[

Nimfa-mama [501]

Using the knowledge of computational language in C++ it is possible to write a code that given an array of integers a, your task is to count the number of pairs i and j.

<h3>Writting the code:</h3>

// C++ program for the above approach

#include <bits/stdc++.h>

using namespace std;

// Function to find the count required pairs

void getPairs(int arr[], int N, int K)

{

// Stores count of pairs

int count = 0;

// Traverse the array

for (int i = 0; i < N; i++) {

for (int j = i + 1; j < N; j++) {

// Check if the condition

// is satisfied or not

if (arr[i] > K * arr[j])

count++;

}

cout << count;

}

// Driver Code

int main()

{

int arr[] = { 5, 6, 2, 5 };

int N = sizeof(arr) / sizeof(arr[0]);

int K = 2;

// Function Call

getPairs(arr, N, K);

return 0;

}

See more about C++ code at brainly.com/question/17544466

#SPJ4

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1 year ago