Answer:
<h3>Rules for Naming Variables</h3><h3>The first character must be a letter or an underscore (_). You can't use a number as the first character. The rest of the variable name can include any letter, any number, or the underscore. You can't use any other characters, including spaces, symbols, and punctuation marks.</h3>
<em><u>#</u></em><em><u>M</u></em><em><u>a</u></em><em><u>r</u></em><em><u>k</u></em><em><u>a</u></em><em><u>s</u></em><em><u>b</u></em><em><u>r</u></em><em><u>a</u></em><em><u>i</u></em><em><u>n</u></em><em><u>l</u></em><em><u>e</u></em><em><u>s</u></em><em><u>s</u></em><em><u>p</u></em><em><u>l</u></em><em><u>e</u></em><em><u>a</u></em><em><u>s</u></em><em><u>e</u></em><em><u>✅</u></em>
A software that does and allows this is free software
Answer:
The answer to this question is given below in the explanation section. However, the correct option is B. i.e a for loop.
Explanation:
The correct answer to this question is for-loop. Beause, in the for loop is written to perform a certain number of iterations or to iterate until a specific result is achieved.
for example, to calculate the average of 10 numbers, we use the for loop to iterate statements written in its body 10 times to calculate the average.
when the specific numbers of the iteration are reached it will get terminated.
for example
for (int i=0; i<=10; i++)
{
// perfrom average, code
}
this loop iterate 10 times or less, when it reached above the limit it gets stopped.
however, the other options are not correct, because an ordered data structure is like a list of order data, print function print something to the user or on screen. while an array is like a list of data of the same data type.
This question is incomplete. The complete question is given below:
Your company has just brought a new 22-core processor, and you have been tasked with optimizing your software for this processor. You will run four applications on this system, but the resource requirements are not equal. Assume the system and application characteristics listed in table 1.1
Table 1.1 Four Applications
Application
A
B
C
D
% resources needed
41
27
18
14
% parallelizable
50
80
60
90
The percentage of resources of assuming they are all run in serial. Assume that when you parallelize a portion of the program by X, the speedup for that portion is X.
a. How much speedup would result from running application A on the entire 22-core processor, as compared to running it serially?
Answer:
Speedup = 50
Explanation:
- The speedup for that portion is x if we parallelize a portion of that program by X.
- If the whole program has no parallelize portion or in other words the whole program is running serially then the speedup will be zero.
- So in this scenario if parallelizable portion of A is 50% so according to above description the speedup is 50.
