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Artyom0805 [142]
2 years ago
9

Assuming P != NP, which of the following is true ?

Computers and Technology
2 answers:
7nadin3 [17]2 years ago
8 0

Answer:

B. NP-complete \cap P = \Phi

Explanation:

The is because no NP-Complete problem can be solved in polynomial time.

Assuming one NP-Complete problem can be solved in polynomial time, then all NP problems can solved in polynomial time. If that is the case, then NP and P set become same which contradicts the given condition.

White raven [17]2 years ago
7 0

Answer:

Answer is B, see explanations.

Explanation:

Answer is (B) NP−complete∩P=ϕ

Since, P≠NP, there is at least one problem in NP, which is harder than all P problems. Lets take the hardest such problem, say X. Since, P≠NP,X∉P .

Now, by definition, NP−complete problems are the hardest problems in NP and so X problem is in NP−complete. And being in NP, X can be reduced to all problems in NP−complete, making any other NP−complete problem as hard as X. So, since X∉P, none of the other NP−complete problems also cannot be in P.

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As the performance of PCs steadily improves, computers that in the past were classified as midrange computers are now marketed a
taurus [48]

Answer: PC servers

Explanation: PC(Personal computer) servers are the helps in the utilization of the network services to the software and hardware of the computer. It manages the network resources . In the past time ,Mindrange computers, mini computers etc are present time's PC servers.

Other options are incorrect because laptops ,PDA(Personal digital assistant) or tablets are the modern version of the computer units that are portable and compact with fast processing. Thus the correct option is option PC servers.

5 0
3 years ago
Prewritten, commercially available sets of software programs that eliminate the need for a firm to write its own software progra
Irina18 [472]

Answer:

Option A i.e., software package.

Explanation:

Prewritten, available on the market the software packages, which not necessitate a company's desire to actually compose its software for certain functionalities, will be applied to as software packages.

A software package is essentially a finite number of programs or software components that function together so that they achieve different tasks and objectives.

5 0
3 years ago
Consider the following method, inCommon, which takes two Integer ArrayList parameters. The method returns true if the same integ
sp2606 [1]

Answer:

The answer is "Option b".

Explanation:

In the given code, a static method "inCommon" is declared, that accepts two array lists in its parameter, and inside the method two for loop is used, in which a conditional statement used, that checks element of array list a is equal to the element of array list b. If the condition is true it will return the value true, and if the condition is not true, it will return a false value. In this, the second loop is not used because j>0 so will never check the element of the first element.

6 0
3 years ago
Apakah ada yang bisa menjelaskan potongan source code ini?
GarryVolchara [31]

This code attempts to fuse two strings together. So,

fuse("Apple", "Banana")

would return "ABpapnlaen a"

However, there are a couple of things wrong with this code:

- The for loop is incomplete (probably a copy paste error)

- It is unclear from the code if the array jawaban will overflow if kata1 and kata2 are large (it probably will)

- Biggest problem: the jawaban array is declared on the stack, which means it will be cleaned up when the function returns. So the caller of this function will reference unallocated memory! This is a huge bug!!

6 0
3 years ago
Find the propagation delay for a signal traversing the in a metropolitan area through 200 km, at the speed of light in cable (2.
Ede4ka [16]

Answer:

t= 8.7*10⁻⁴ sec.

Explanation:

If the signal were able to traverse this distance at an infinite speed, the propagation delay would be zero.

As this is not possible, (the maximum speed of interactions in the universe is equal to the speed of light), there will be a finite propagation delay.

Assuming that the signal propagates at a constant speed, which is equal to 2.3*10⁸ m/s (due to the characteristics of the cable, it is not the same as if it were propagating in vaccum, at 3.0*10⁸ m/s), the time taken to the signal to traverse the 200 km, which is equal to the propagation delay, can be found applying the average velocity definition:

v = \frac{(xf-xo)}{(t-to)}

If we choose x₀ = 0 and t₀ =0, and replace v= 2.3*10⁸ m/s, and xf=2*10⁵ m, we can solve for t:

t =\frac{xf}{v}  =\frac{2e5 m}{2.3e8 m/s} =8.7e-4 sec.

⇒ t = 8.7*10⁻⁴ sec.

4 0
3 years ago
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