Why hasn't net neutrality been taken away? It's past Dec 14.

Write a program that asks the user to input an integer named numDoubles. Create a dynamic array that can store numDoubles double

Hitman42 [59]

Answer:

Answered in C++

#include <iostream>

using namespace std;

int main() {

int size;

cout<<"Length of array: ";

cin >> size;

double *numDoubles = new double[size];

double sum =0;

cout<<"Array Elements: ";

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

cin>>numDoubles[i];

sum += numDoubles[i];

}

delete [] numDoubles;

cout<<"Average: "<<sum/size;

return 0;

}

Explanation:

This line declares the size (or length) of the array

int size;

This line prompts the user for the array length

cout<<"Length of array: ";

This gets the input from the user

cin >> size;

This dynamically declares the array as of double datatype

double *numDoubles = new double[size];

This declares and initializes sum to 0

double sum =0;

This prompts the user for elements of the array

cout<<"Array Elements: ";

The following iteration gets input from the user and also calculates the sum of the array elements

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

cin>>numDoubles[i];

sum += numDoubles[i];

}

This deletes the array

delete [] myarray;

This calculates and prints the average of the array

cout<<"Average: "<<sum/size;

6 0

3 years ago

AN OBSTACLE OR BARRIER THAT MAY PREVENT YOU FROM ACCOMPLISHING YOUR GOAL IS:

marishachu [46]

Depends on what kind of goal.
If I gave a goal of eating 8 hot dogs then a barrier would be a physical will of your body to not eat more. It would be a limit your body can take so it’s different from a mental or other barrier.
If you wanted to cross an area of land but there is a fence in the way then that’s another type of barrier.
Or there are mental barriers. If you wanted to ask someone out to prom, you may be stopped by a mental barrier of self hatred. Feeling like you’re not good enough.

Hopefully this helps!

5 0

3 years ago

Rupa would like to quickly insert a table into her document without having to worry about formatting the data in the table. Whic

USPshnik [31]

Internet tap and log on application

6 0

2 years ago

Read 2 more answers

An enterprise DBMS is automatically capable of serving as a mobile DBMS. There are no special issues raised by mobility. True Fa

pishuonlain [190]

Answer:

The answer is True.

Explanation:

The enterprise DBMS can also be used with mobile DBMS.

Enterprise DMBS is the latest version of DBMS which is used in organizations and enterprises to handle a huge amount of Data.

Enterprise Database Management System is mainly designed to do large work simultaneously. It can handle multiple queries simultaneously.

Multiple users (about 100-10,000 users) can access data at the same time and even they can manipulate simultaneously.

The features of Enterprise DBMS is to work efficiently, multi processing, fast, accurately, and handling huge burden of data.

4 0

2 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

2 years ago