Answer:
Check the explanation
Explanation:
in order to make sure that the built into the new systems, the below input and output levels of security control have to be considered for personal trainers staff.
There are:
Network Security
Application security
Physical security
User security
File security
Procedural security
These securities levels are consistent and advantage in building the decisions regarding system security.
There are the following policies applied to each control:
Network Security: Network interface and traffic controlling, encryption of data.
Application Security: Verification of data input/output and update of software proceedings.
Physical Security: Security to servers, computers and physical environment.
User Security: Protection passwords and identify management skills.
File Security: Access Permission allocated to users and editing authorizations given to users.
Procedural Security: Certify safety protection to perform decision-making activities and document shredders.
Finally, consider another main point is depends the new system performance are system response time, throughput, bandwidth time and turnaround time.
Answer:
The comparison is based on memory organization schemes of contiguous memory allocation and paging with respect to External fragmentation, Internal fragmentation and Ability to share code across processes.
Explanation:
Memory organization schemes of contiguous memory allocation:
Contiguous memory allocation schemes suffers from external fragmentation. The reason is that address space is distributed contiguously and the holes and gaps keep growing when the old processes die and new processes are introduced. The variable size partition suffers from external fragmentation however the fixed size partitions do not suffer from external fragmentation. Contiguous memory allocation with variable size partitions does not encounter an internal fragmentation but with fixed size partitions suffers from internal fragmentation. Contiguous memory allocation does not support sharing code across processes. This is because the virtual memory segment of a process is not fragmented into non-contiguous fine grained blocks.
Paging:
Paging does not encounter external fragmentation as pages are of the fixed or equal size. So this reduces external fragmentation. However paging suffers from internal fragmentation. This is because a process can request more space or it can request for a less space. When page is allocated to the such a process that page is no longer utilized.This results in internal fragmentation because of the wastage of space even when the page has internal space but cannot be fully utilized. Paging allows to share code across processes.
Answer:
-26
Explanation:
The given binary number is 1110 0101. Also given that the signed binary number is represented using one's compliment.
We begin by computing the 1s complement representation of 1110 0101 by inverting the bits: 00011010
Converting 00011010 to decimal, it corresponds to 26.
So the 1s complement of the original number is 26. This means that the original number was -26.
Answer:
I don't know if this is right output is {1,3}
Explanation:
Answer:
Here is the script:
function dd = functionDMS(dd)
prompt= 'Enter angle in DD form ';
dd = input(prompt)
while (~checknum(dd))
if ~checknum(dd)
error('Enter valid input ');
end
dd = input(prompt)
end
degrees = int(dd)
minutes = int(dd - degrees)
seconds = ( dd - degrees - minutes / 60 ) * 3600
print degrees
print minutes
print seconds
print dd
Explanation:
The script prompts the user to enter an angle in decimal degree (DD) form. Next it stores that input in dd. The while loop condition checks that input is in valid form. If the input is not valid then it displays the message: Enter valid input. If the input is valid then the program converts the input dd into degrees, minutes and seconds form. In order to compute degrees the whole number part of input value dd is used. In order to compute the minutes, the value of degrees is subtracted from value of dd. The other way is to multiply remaining decimal by 60 and then use whole number part of the answer as minutes. In order to compute seconds subtract dd , degrees and minutes values and divide the answer by 60 and multiply the entire result with 3600. At the end the values of degrees minutes and seconds are printed. In MATLAB there is also a function used to convert decimal degrees to degrees minutes and seconds representation. This function is degrees2dms.
Another method to convert dd into dms is:
data = "Enter value of dd"
dd = input(data)
degrees = fix(dd);
minutes = dd - degrees;
seconds = (dd-degrees-minutes/60) *3600;