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3 years ago

What is an avenue in a Karel world?

Computers and Technology

2 answers:

melamori03 [73]3 years ago

6 0

Answer:

Column

hope it helps:)

Andrews [41]3 years ago

3 0

Answer:

Karel's World contains horizontal streets (running east-west) and vertical avenues (running north-south). can contain interior wall sections that are positioned between adjacent street corners and prevent Karel from moving directly between these street corners.

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Describe the benefits and drawback of using virtualisation on a single machine and also in a cluster environment

Sedbober [7]

Answer:

Virtualization is the technology that is used to create virtual representation and version of the different devices like software, many storage devices and server.

It also help in organize the working framework where the structure isolates the asset into at least one execution situations.

The benefits of using the virtualization on the single machine are:

The virtualization increased the productivity when it used in the cluster environment.
We can able to access the different resources faster.
The data center management is simple while using the virtualization in the cluster environment.

The drawback of using virtualization on single machine are:

The virtual machine is less efficient as compared to the real machines.
The virtual machine indirectly access the computer hardware system so that is why it is less efficient and also consume more power to run the system.

5 0

3 years ago

To save a presentation, tap or click the save button on the Blank

frozen [14]

If this is a true or false statement then the answer is true unless you have it on automatic save

4 0

3 years ago

An application programming interface (API) is ________. Select one: A. the code to interface between an application and RAM B. t

Olegator [25]

Answer:

An application programming interface (API) is the code the CPU recognizes to perform a procedure in an application.

Explanation:

An application programming interface (API) is the code the CPU recognizes to perform a procedure in an application. API allows an application to communicate with another application, or an operating system, database, network, etc.

An Application Programming Interface (API) creates a consideration for a problem and specifies how clients should interact with software components that implement a solution to that problem.

More recently, API has been used to refer to a specific type of interface between a client and a server, which has been described as a “contract” between both - such that if the client makes a request in a specific format, it will always get a response in a specific format or initiate a defined action.This is a specialized form of API, defined as a Web API.

3 0

3 years ago

Referat noaptea la școala cu sfârșit de groază

solniwko [45]

Te ajut eu daca vrei tu la refera

7 0

2 years ago

2. Consider a 2 GHz processor where two important programs, A and B, take one second each to execute. Each program has a CPI of

allsm [11]

Answer:

program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.

Explanation:

Finding number of instructions in A and B using time taken by the original processor :

The clock speed of the original processor is 2 GHz.

which means each clock takes, 1/clockspeed

= 1 / 2GH = 0.5ns

Now, the CPI for this processor is 1.5 for both programs A and B. therefore each instruction takes 1.5 clock cycles.

Let's say there are n instructions in each program.

therefore time taken to execute n instructions

= n * CPI * cycletime = n * 1.5 * 0.5ns

from the question, each program takes 1 sec to execute in the original processor.

therefore

n * 1.5 * 0.5ns = 1sec

n = 1.3333 * 10^9

So, number of instructions in each program is 1.3333 * 10^9

the new processor :

The cycle time for the new processor is 0.6 ns.

Time taken by program A = time taken to execute n instructions

= n * CPI * cycletime

= 1.3333 * 10^9 * 1.1 * 0.6ns

= 0.88 sec

Time taken by program B = time taken to execute n instructions

= n * CPI * cycletime

= 1.3333 * 10^9 * 1.4 * 0.6

= 1.12 sec

Now, program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.

5 0

3 years ago