Answer:
Option b outFile.open("outputData.out");
Explanation:
In C++, there are several classes given to handle output and input of characters to or from files. They are:
- ofstream that write on files
- ifstream that read from files
We can use an object of ofstream to hold the contents that we wish to output to an external file. The general syntax is as follows:
ofstream_obj.open(file_name)
This will create a file with a specific file name and it possesses all the contents from the obstream_obj.
Answer:
import random
randomlist = []
for i in range(0,20):
n = random.randint(-29,30)
if n < 0 :
n = 100
randomlist.append(n)
print(randomlist)
Explanation:
The random module is first imported as it takes care of random. Number generation.
An empty list called randomliay is created to hold the generated random integers.
Using a for loop, we specify the range of random numbers we want.
Inside the for loop ; we attach our generated random integer which will be in the range (-29 to 30) in a variable n
For each n value generated, if the value is less than 0( it is negative, since all the values are integers), replace the value with 100.
Answer:
a) Speedup gain is 1.428 times.
b) Speedup gain is 1.81 times.
Explanation:
in order to calculate the speedup again of an application that has a 60 percent parallel component using Anklahls Law is speedup which state that:
Where S is the portion of the application that must be performed serially, and N is the number of processing cores.
(a) For N = 2 processing cores, and a 60%, then S = 40% or 0.4
Thus, the speedup is:
Speedup gain is 1.428 times.
(b) For N = 4 processing cores and a 60%, then S = 40% or 0.4
Thus, the speedup is:
Speedup gain is 1.81 times.
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Answer:
The correct option is A
Explanation:
In project management, earliest finish time for activity A refers to the earliest start time for succeeding activities such as B and C to start.
Assume that activities A and B comes before C, the earliest finish time for C can be arrived at by computing the earliest start-finish (critical path) of the activity with the largest EF.
That is, if two activities (A and B) come before activity C, one can estimate how long it's going to take to complete activity C if ones knows how long activity B will take (being the activity with the largest earliest finish time).
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