Answer:
C. bachelor’s degree in filmmaking
Explanation:
Because Andy wants to become a multimedia producer, the degree that would best help him achieve his goal is a bachelor's degree in filmmaking.
Multimedia has to do with both audio, video and graphics or animations because it encompasses multiple media files.
Filmmaking has to do with the various forms of making and producing films. Filmmaking has to do with multiple media (multimedia) and it would help him achieve his goal.
Answer: Even though the hardware is inexpensive the writing of program is not efficient through this method as proper development of program is necessary for the clear execution due to factors like:-
- The facility of writing program even the cost of hardware is less but it is not a free facility.
- It also has a slower processing for the execution of the program
- The construction of the efficient program is necessary for the compilation and execution of it rather than poorly constructed program is worthless and inefficient in working.
Answer:
#include <iostream>
using namespace std;
int main () {
// for loop execution
int semester_fees=8000;
int b=1;
cout<<"there are 5 years or 10 semester fees breakdown is given below"<<endl;
for( int a = 1; a <=5; a++ ) {
semester_fees = semester_fees*1.03;
cout<<"Semester fees for each of semester"<<b++<<"and"<<b++<<"is:$"<<semester_fees<<endl;
}
return 0;
}
Explanation:
code is in c++ language
Fees is incremented yearly and i have considered only two semester per year
Answer:
So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %
Explanation:
Previous concepts
Input/output operations per second (IOPS, pronounced eye-ops) "is an input/output performance measurement used to characterize computer storage devices like hard disk drives (HDD)"
Solution to the problem
For this case since we have 4GB, but 512 MB are destinated to the operating system, we can begin finding the available RAM like this:
Available = 4096 MB - 512 MB = 3584 MB
Now we can find the maximum simultaneous process than can use with this:
And then we can find the maximum wait I/O that can be tolerated with the following formula:
The expeonent for p = 14 since we got 14 simultaneous processes, and the rate for this case would be 99% or 0.99, if we solve for p we got:
So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %
