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Anna11 [10]
3 years ago
11

Which of these can be represented visually by a flowchart?

Computers and Technology
2 answers:
kifflom [539]3 years ago
5 0
I think it is c if not sorry
Mariana [72]3 years ago
4 0

Answer:

i think its C.

if its wrong i'm truly sorry

Explanation:

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Consider the following sequence, defined from n=2 to 15 (inclusive). Pn=n2−1. Produce a list named primes which only contains va
omeli [17]

Answer:

primes = []

for n in range(2,16):

   pn = n*n - 1

   if is_prime(pn):

       primes.append(pn)

5 0
3 years ago
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GalinKa [24]
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3 years ago
The default case is required in the switch selection statement.
babunello [35]

Answer: False

Explanation:

 The given statement is false, as it is not compulsory that the default case is require in switch selection statement. If the default case are not properly specified in the statement then, there will be no execution occur in the switch selection statement.

It is sometimes good to have default case in switch selection statement, but it is not mandatory. The default case only executed when the correct case is present and none of the case matches in the given statement.

5 0
3 years ago
Which feature of a blog helps to store and retrieve older posts?
Rashid [163]
Could it be archive posts? I'm not sure, but I believe it's archive posts.
8 0
3 years ago
Read 2 more answers
Use semaphore(s) to solve the following problem. There are three processes: P1, P2, and P3. Each process Pi has a segment of cod
Lisa [10]

Answer:

See explaination

Explanation:

Here we will use two semaphore variables to satisfy our goal

We will initialize s1=1 and s2=1 globally and they are accessed by all 3 processes and use up and down operations in following way

Code:-

s1,s2=1

P1 P2 P3

P(s1)

P(s2)

C1

V(s2) .

P(s2). .

. C2

V(s1) .

P(s1)

. . C3

V(s2)

Explanation:-

The P(s1) stands for down operation for semaphore s1 and V(s1) stands for Up operation for semaphore s1.

The Down operation on s1=1 will make it s1=0 and our process will execute ,and down on s1=0 will block the process

The Up operation on s1=0 will unblock the process and on s1=1 will be normal execution of process

Now in the above code:

1)If C1 is executed first then it means down on s1,s2 will make it zero and up on s2 will make it 1, so in that case C3 cannot execute because P3 has down operation on s1 before C3 ,so C2 will execute by performing down on s2 and after that Up on s1 will be done by P2 and then C3 can execute

So our first condition gets satisfied

2)If C1 is not executed earlier means:-

a)If C2 is executed by performing down on S2 then s2=0,so definitely C3 will be executed because down(s2) in case of C1 will block the process P1 and after C3 execute Up operation on s2 ,C1 can execute because P1 gets unblocked .

b)If C3 is executed by performing down on s1 then s1=0 ,so definitely C2 will be executed now ,because down on s1 will block the process P1 and after that P2 will perform up on s1 ,so P1 gets unblocked

So C1 will be executed after C2 and C3 ,hence our 2nd condition satisfied.

4 0
3 years ago
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