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DiKsa [7]
3 years ago
7

How long will it take to transfer 1GB data on USB 2.0, on USB 3.0

Computers and Technology
1 answer:
cupoosta [38]3 years ago
8 0

Answer:

Have a great day, Here is the answer to your question:

It will take around 18 seconds

Explanation:

So in principle 1GB of data to be uploaded running at peak speed is suggested to take around 18 seconds but in fact, DCD test results show USB 2.0 takes 3 minutes 18 seconds to complete a 1GB switch. Whereas USB 3.0 can accommodate up to 5gbps of data transferred-more than 10 times faster than its predecessor.

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What will be the output of the following query on the table below?
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Answer:

b

Explanation:

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Data with values that change continuously or smoothly over time is known as:
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Answer:

A.  \:  \boxed{analog  \: data}

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A milk carton can hold 3.78 liters of milk. Each morning, a dairy farm ships cartons of milk to a local grocery store. The cost
makvit [3.9K]

Answer:

milk_produced = float(input("Enter the total amount of milk produced in the morning: "))

liter_cost = float(input("Enter the cost of producing one liter of milk: "))

carton_profit = float(input("Enter the profit on each carton of milk: "))

carton_needed = round(milk_produced / 3.78)

cost = milk_produced * liter_cost

profit = carton_profit * carton_needed

print("The number of milk cartons needed to hold milk is " + str(carton_needed))

print("The cost of producing milk is " + str(cost))

print("The profit for producing milk is " + str(profit))

Explanation:

*The code is in Python.

Ask the user to enter milk_produced, liter_cost and carton_profit

Calculate the number of milk cartons needed, divide the milk_produced by the capacity (3.78) of a cartoon and round the result

Calculate the cost, multiply the milk_produced by liter_cost

Calculate the profit, multiply the carton_profit by carton_needed

Print the results

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3 years ago
You are asked to simulate a binary search algorithm on an array of random values.An array is the list of similar type of element
Alex Ar [27]

Answer:

Explanation:

Problem statement:

to simulate a binary search algorithm on an array of random values.

Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.

Input/output description

Input:

Size of array: 4

Enter array:10  20 30 40

Enter element to be searched:40

The Output will look like this:

Element is present at index 3

Algorithm and Flowchart:

We basically ignore half of the elements just after one comparison.

Compare x with the middle element.

If x matches with middle element, we return the mid index.

Else If x is greater than the mid element, then x can only lie in right half subarray after the mid element. So we recur for right half.

Else (x is smaller) recur for the left half.

The Flowchart can be seen in the first attached image below:

Program listing:

// C++ program to implement recursive Binary Search

#include <bits/stdc++.h>

using namespace std;

// A recursive binary search function. It returns

// location of x in given array arr[l..r] is present,

// otherwise -1

int binarySearch(int arr[], int l, int r, int x)

{

   if (r >= l) {

       int mid = l + (r - l) / 2;

       // If the element is present at the middle

       // itself

       if (arr[mid] == x)

           return mid;

       // If element is smaller than mid, then

       // it can only be present in left subarray

       if (arr[mid] > x)

           return binarySearch(arr, l, mid - 1, x);

       // Else the element can only be present

       // in right subarray

       return binarySearch(arr, mid + 1, r, x);

   }

   // We reach here when element is not

   // present in array

   return -1;

}

int main(void)

{ int n,x;

cout<<"Size of array:\n";

cin >> n;

int arr[n];

cout<<"Enter array:\n";

for (int i = 0; i < n; ++i)

{ cin >> arr[i]; }

cout<<"Enter element to be searched:\n";

cin>>x;

int result = binarySearch(arr, 0, n - 1, x);

   (result == -1) ? cout << "Element is not present in array"

                  : cout << "Element is present at index " << result;

   return 0;

}

The Sample test run of the program can be seen in the second attached image below.

Time(sec) :

0

Memory(MB) :

3.3752604177856

The Output:

Size of array:4

Enter array:10  20 30 40

Enter element to be searched:40

Element is present at index 3

Conclusions:

Time Complexity:

The time complexity of Binary Search can be written as

T(n) = T(n/2) + c  

The above recurrence can be solved either using Recurrence T ree method or Master method. It falls in case II of Master Method and solution of the recurrence is Theta(Logn).

Auxiliary Space: O(1) in case of iterative implementation. In case of recursive implementation, O(Logn) recursion call stack space.

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A web designers favorite snack is cookies.
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