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4 years ago

Explain the difference between an i/o‐bound process and a cpu‐bound process.

1 answer:

7 0

An I/O bound process can't finish because it's waiting for I/O (which is provided by the OS). A CPU bound process can't finish because it can't get scheduled in the processor enough.

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Jennifer has to set up a network in a factory with an environment that has a lot of electrical interference. Which cable would s

jeka57 [31]

The answer is STP (Shielded Twisted Pair) cables .

Compared to Unshielded Twisted Pair, STP offers additional safeguards against Electromagnetic interference (EMI) and crosstalk. While UTP reduces some EMI, STP cables effectively block interference and ensure high-speed performance. STP cables have a metallic foil that cancels out electromagnetic interference. They are the preferred cables that protect against high-level EMI from, power lines, radar systems, or electromagnetic fields .

6 0

3 years ago

The running time of Algorithm A is (1/4) n2+ 1300, and the running time of another Algorithm B for solving the same problem is 1

Mnenie [13.5K]

Answer:

Answer is explained below

Explanation:

The running time is measured in terms of complexity classes generally expressed in an upper bound notation called the big-Oh ( "O" ) notation. We need to find the upper bound to the running time of both the algorithms and then we may compare the worst case complexities, it is also important to note that the complexity analysis holds true (and valid) for large input sizes, so, for inputs with smaller sizes, an algorithm with higher complexity class may outperform the one with lower complexity class i.e, efficiency of an algorithm may vary in cases where input sizes are smaller & more efficient algorithm might be outperformed by the lesser efficient algorithms in those cases.

That's the reason why we consider inputs of larger sizes when comparing the complexity classes of the respective algorithms under consideration.

Now coming to our question for algorithm A, we have,

let F(n) = 1/4x² + 1300

So, we can tell the upper bound to the function O(F(x)) = g(x) = x2

Also for algorithm B, we have,

let F(x) = 112x - 8

So, we can tell the upper bound to the function O(F(x)) = g(x) = x

Clearly, algorithmic complexity of algorithm A > algorithmic complexity of algorithm B

Hence we can say that for sufficiently large inputs , algorithm B will be a better choice.

Now to find the exact location of the graph in which algorithmic complexity for algorithm B becomes lesser than

algorithm A.

We need to find the intersection point of the given two equations by solving them:

We have the 2 equations as follows:

y = F(x) = 1/4x² + 1300 __(1)

y = F(X) = 112x - 8 __(2)

Let's put the value of from (2) in (1)

=> 112x - 8 = 1/4x² + 1300

=> 112x - 0.25x² = 1308

=> 0.25x² - 112x + 1308 = 0

Solving, we have

=> x = (112 ± 106) / 0.5

=> x = 436, 12

We can obtain the value for y by putting x in any of the equation:

At x=12 , y= 1336

At x = 436 , y = 48824

So we have two intersections at point (12,1336) & (436, 48824)

So before first intersection, the

Function F(x) = 112x - 8 takes lower value before x=12

& F(x) = 1/4x² + 1300 takes lower value between (12, 436)

& F(x) = 112x - 8 again takes lower value after (436,∞)

Hence,

We should choose Algorithm B for input sizes lesser than 12

& Algorithm A for input sizes between (12,436)

& Algorithm B for input sizes greater than (436,∞)

8 0

3 years ago

Unit testing: Select one:

ser-zykov [4K]

Answer:

e. tests each program separately.

Explanation:

Unit testing -

It is one of the software testing where the individual units are tested , is referred to as unit testing.

The focus of this step is to scan each and every unit separately and thoroughly so as to avoid any type of damage or malfunctioning .

Hence, from the question, the correct statement for unit testing is e. tests each program separately.

5 0

3 years ago

4. When the keyword ____________________ is followed by a parenthesis, it indicates a call to the superclass constructor.

Andreyy89

Answer:

Super

Explanation:

In object-oriented programming (OOP) language, an object class represents the superclass of every other classes when using a programming language such as Java. The superclass is more or less like a general class in an inheritance hierarchy. Thus, a subclass can inherit the variables or methods of the superclass.

Basically, all instance variables that have been used or declared in any superclass would be present in its subclass object.

Hence, when the keyword super is followed by a parenthesis, it indicates a call to the superclass constructor and it should be the first statement declared in the subclass constructor.

5 0

3 years ago

A franchise restaurant is attempting to understand if customers think their service is good day-to-day by summarizing a series o

Crazy boy [7]

Answer:

C++.

Explanation:

int main() {

int num_days_scores;

cout<<"How many days of scores? ";

cin<<num_days_scores;

cout<<endl;

///////////////////////////////////////////////////////////////////////////

// Get scores for each day and add it to array

int score = 0;

int* score_array = new int[num_days_scores];

for (int i = 0; i < num_days_scores; i++) {

cout<<Enter score for day "<<i+1<<": ";

cin<<score;

score_array[i] = score;

}

////////////////////////////////////////////////////////////////////////////

// Calculate total of scores by traversing array

int final_score = 0;

for (int i = 0; i < num_days_scores; i++) {

final_score += score_array[i];

}

cout<<"The total score of the "<<num_days_scores<<" days is "<<final_score;

////////////////////////////////////////////////////////////////////////////

delete[] score_array;

return 0;

}

7 0

3 years ago