<span>Images are available in many formats, such as tif, bmp, gif, jpeg, and png.
</span>Hope that helps.
Answer:
A website is a group of globally accessible, interlinked web pages which have a single domain name. A web application is a software or program which is accessible using any web browser. Developing your website helps you in branding your business.
Answer:
A domestic air carrier airplane lands at an intermediate airport at 1815Z. The latest time it may depart without a specific authorization from an aircraft dispatcher is <u>1915Z (1 hour)</u>.
Explanation:
Under the domestic operations, an airplane landed on intermediate airport can remain their for not more than one hour so the time would me 1 hour.
Here the time represented by 1815Z and 1915Z is in Zulu Time Zone as depicted from letter "Z". The first two digits represent the hour (0-24) and the next two represent the minutes (0-59).
- Here the landing time is 6:15 pm while departing time is after one hour that is 7:15 pm (1915Z).
i hope it will help you.
Answer:
The method is as follows:
double square(int num){
return num*num;
}
Explanation:
Written in C++
This first line defines the method
double square(int num){
This line returns the square of num
return num*num;
}
<em>I've added the full program as an attachment where I include the main method</em>
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
