The level of system and network configuration is required for Cui will be "Moderate confidentiality".
- The data that really should be safeguarded or disseminated under appropriate regulations, rules, and conservative government objectives or guidelines, would be considered as CUI.
- The FISMA demands CUI Fundamental to somehow be safeguarded somewhere at the FISMA Conservative category and may even be labeled as CUI as well as regulated.
Thus the above answer is right.
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My early test cases were greatly aided by the user stories. They gave a thorough summary of what a user can expect from the product as well as data on their past expectations and experiences.
Explanation:
The intricacies of the user's surroundings and the exact technological needs for the product, however, were absent from the user stories.
Greetings, Software Developers!
I'm trying to create more in-depth test cases for my job as a quality test for the online travel software. To do this, I'd like to learn more about the environment of the user and the software's technical requirements. Please give me more details about the technical specs and the user environment.
I'm grateful.
[Name]
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Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed.
(I'm assuming that := is a typo and you mean just =, by the way)
In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.
This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.
HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES.
</span>
As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2).
(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.
A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
The script that Andy would want to use is
Javascript, here is the source code: document.getElementById("para1").innerHTML = formatAMPM();
function formatAMPM() {var d = new Date(), minutes = d.getMinutes().toString().length == 1 ? '0'+d.getMinutes() : d.getMinutes(), hours = d.getHours().toString().length == 1 ? '0'+d.getHours() : d.getHours(), ampm = d.getHours() >= 12 ? 'pm' : 'am', months = ['Jan','Feb','Mar','Apr','May','Jun','Jul','Aug','Sep','Oct','Nov','Dec'], days = ['Sun','Mon','Tue','Wed','Thu','Fri','Sat'];return days[d.getDay()]+' '+months[d.getMonth()]+' '+d.getDate()+' '+d.getFullYear()+' '+hours+':'+minutes+ampm;<span>}
The HTML code needed to call this Javascript on his website is this: </span><span><div id="para1"></div>
</span>
You could call the Javascript up using PHP.
Hope this helps! Good luck! :)
