<span>The following statements are true:
1. Leased lines require little installation and maintenance expertise.
( A high service quality is offered by point-to point system)
2. Leased lines provide highly flexible bandwidth scaling.
(This Allows a Constant availability)</span>
Answer:
answer is "is there an early pay discount"
Explanation:
Answer:
Hi there! This can be implemented in a simple Python function which uses the "random" module to generate the number.
Explanation:
Using Python as the languge, we can write a the below code in a file called styles.py. The first line imports the randint function from the "random" module. The setStyles() function declares an array or 5 elements (here I have just used numbers but these could be string names of the stylesheets as well). Next, styleNum is assigned the random number and the associated stylesheet is selected from the array of stylesheets.
styles.py
from random import randint
def setStyles():
stylesheets = [1,2,3,4,5];
styleNum = randint(1,5);
stylesheet = stylesheets[styleNum];
print(stylesheet);
setStyles();
Answer:
(a)The CPU B should be selected for the new computer as it has a low clock cycle time which implies that it will implement the process or quicker when compared to the CPU A.
(b) The CPU B is faster because it executes the same number of instruction in a lesser time than the CPU A
.
Explanation:
Solution
(a)With regards to the MIPS performance metric the CPU B should be chosen for the new computer as it low clock cycle time which implies that it will implement the process or quicker when compared to the CPU A and when we look at the amount of process done by the system , the CPU B is faster when compared to other CPU and carries out same number of instruction in time.
The metric of response time for CPU B is lower than the CPU A and it has advantage over the other CPU and it has better amount as compared to CPU A, as CPU B is carrying out more execution is particular amount of time.
(b) The execution can be computed as follows:
Clock cycles taken for a program to finish * increased by the clock cycle time = the Clock cycles for a program * Clock cycle time
Thus
CPU A= 5*10^6 * 60*10^-9 →300*10^-3 →0.3 second (1 nano seconds =10^-9 second)
CPU B= 3 *10^6 * 75*10^-9 → 225*10^-3 → 0.225 second
Therefore,The CPU B is faster as it is executing the same number of instruction in a lesser time than the CPU A
