Answer:
Input, output, storage.
Explanation:
Input is used to interact with, or send data to the computer (mouse, keyboards, etc.). Output provides output to the user from the computer (monitors, printers, etc.). And storage which stores data processed by the computer (hard drives, flash drives, etc.).
Answer:
In studies of technology and innovation, it is common to distinguish between different levels of technology. A common distinction is between technologies as processes or products and technology systems. Another distinction is between technology s hardware, software and orgware; sometimes socioware is added as yet another layer or category. In he studies, the focus is usually on a particular technology, either a process technology or product
Ž echnology with product technology broadly de In studies of technology and innovation, it is
common to distinguish between different levels of technology. A common distinction is between technologies as processes or products and technology. Another distinction is between technology as hardware, software and orgware; sometimes socioware is added as yet another layer or category. In the studies, the focus is usually on a particular technology, either a process technology or product.
Explanation:
Answer:
padna jana ta yarrr kasto this yarr
Answer:
cout << setprecision(2)<< fixed << number;
Explanation:
The above statement returns 12.35 as output
Though, the statement can be split to multiple statements; but the question requires the use of a cout statement.
The statement starts by setting precision to 2 using setprecision(2)
This is immediately followed by the fixed manipulator;
The essence of the fixed manipulator is to ensure that the number returns 2 digits after the decimal point;
Using only setprecision(2) in the cout statement will on return the 2 digits (12) before the decimal point.
The fixed manipulator is then followed by the variable to be printed.
See code snippet below
<em>#include <iostream> </em>
<em>#include <iomanip>
</em>
<em>using namespace std; </em>
<em>int main() </em>
<em>{ </em>
<em> // Initializing the double value</em>
<em> double number = 12.3456; </em>
<em> //Print result</em>
<em> cout << setprecision(2)<< fixed << number; </em>
<em> return 0; </em>
<em>} </em>
<em />
Answer:
Answer is provided in the explanation section
Explanation:
Given data:
Bandwidth of link = 10* 106 bps
Length of packet = 12* 103 bits
Distance of link = 40 * 103m
Transmission Speed = 3 * 108 meters per second
Formulas:
Transmission Delay = data size / bandwidth = (L /B) second
Propagation Delay = distance/transmission speed = d/s
Solution:
Transmission Delay = (12* 103 bits) / (10* 106 bps) = 0.0012 s = 1.2 millisecond
Propagation Delay = (40 * 103 meters)/ (3 * 108mps) = 0.000133 = 0.13 millisecond
