All categories

2 years ago

In a three-tier architecture, the component that runs the program code and enforces the business processes is the:_______.

Computers and Technology

1 answer:

madam [21]2 years ago

8 0

Answer:

Application Server

Explanation:

The Application Server is a component in computer engineering that presents the application logic layer in a three-tier architecture.

This functionality allows client components to connect with data resources and legacy applications.

In this process of interaction, the Application Server runs the program code from Tier 1 - Presentation, through Tier 2 - Business Logic to Tier 3 - Resources, by forcing through the business processes.

You might be interested in

Considering all of the time and effort that is put into creating an online business, it is in a merchant’s best interest to make

givi [52]

Search engine optimization will help increase the number of clicks on the business' website and help to drive the sales by targeting the appropriate audience for buying the products.

<h3>What is search engine optimization?</h3>

Search engine optimization, or SEO, is one of the most important functions of the generation of organic clicks on the website for an online ecommerce business.

It helps in creating a brand awareness, the costs of advertising are much lower, and also helps in improving the experience of the customers of ecommerce site.

Hence, the uses of search engine optimization in online business are as aforementioned.

Learn more about Search Engine Optimization here:

brainly.com/question/14850803

#SPJ1

4 0

1 year ago

Help!!!!!!<br>please<br>Who will give the answer I'll mark him/her as brainlist..

damaskus [11]

17, 12.5, 6, and 10. HOPE THIS HELPS

4 0

2 years ago

Cybercrime has a serious economic impact on national and global economies that costs the _____ of its GDP every year.

maks197457 [2]

Answer:

World at least 1 percent

Explanation:

Cybercrime are crimes that focuses on networks of computers whereby a computer is used to commit the crime or a computer is the target of the crime, with the possible effect of a threat to the security and financial wellbeing of a person, an establishment or country

According to the report by Zhanna Malekos Smith and Eugenia Lostri, 'The Hidden Costs of Cybercrime' the monetary cost of cybercrime which is estimated at $945 billion is <u>a little above 1% of the World's GDP</u> which is around $87.55 trillion.

4 0

2 years ago

Lets assume we are writing a system to backup a list of transactions: class Transaction 1 String TransactioniD: Date TranactionT

zhannawk [14.2K]

Answer:

The answer is "Option d"

Explanation:

In this question, the easiest way that will save the payment on your database in such a process ID-sorting list would be to mark a payment, that's been recorded mostly on the database whenever this payment became used serial number is not transaction ID, and the wrong choice can be defined as follows:

In choice a, It is wrong because it may be processed, however, payments aren't entered through our process, which does not help remove older.
In choice b, the unordered list would not enable any transaction to only be retrieved, that's why it is wrong.
In choice c, it will not be helpful because the includes video is either begin or complete the payment, it will not be helpful to hold it with transaction time.
In choice e, this approach won't help to identify the payments since one date will have a lot of payments over a certain account.

6 0

3 years ago

A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, an

yulyashka [42]

The question is incomplete. It can be found in search engines. However, kindly find the complete question below:

Question

Cites as a pitfall the utilization of a subset of the performance equation as a performance metric. To illustrate this, consider the following two processors. P1 has a clock rate of 4 GHz, average CPI of 0.9, and requires the execution of 5.0E9 instructions. P2 has a clock rate of 3 GHz, an average CPI of 0.75, and requires the execution of 1.0E9 instructions. 1. One usual fallacy is to consider the computer with the largest clock rate as having the largest performance. Check if this is true for P1 and P2. 2. Another fallacy is to consider that the processor executing the largest number of instructions will need a larger CPU time. Considering that processor P1 is executing a sequence of 1.0E9 instructions and that the CPI of processors P1 and P2 do not change, determine the number of instructions that P2 can execute in the same time that P1 needs to execute 1.0E9 instructions. 3. A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, and consider that the processor with the largest MIPS has the largest performance. Check if this is true for P1 and P2. 4. Another common performance figure is MFLOPS (millions of floating-point operations per second), defined as MFLOPS = No. FP operations / (execution time x 1E6) but this figure has the same problems as MIPS. Assume that 40% of the instructions executed on both P1 and P2 are floating-point instructions. Find the MFLOPS figures for the programs.

Answer:

(1) We will use the formula:

CPU time = number of instructions x CPI / Clock rate

So, using the 1 Ghz = 10⁹ Hz, we get that

CPU time₁ = 5 x 10⁹ x 0.9 / 4 Gh

= 4.5 x 10⁹ / 4 x 10⁹Hz = 1.125 s

and,

CPU time₂ = 1 x 10⁹ x 0.75 / 3 Ghz

= 0.75 x 10⁹ / 3 x 10⁹ Hz = 0.25 s

So, P2 is actually a lot faster than P1 since CPU₂ is less than CPU₁

(2)

Find the CPU time of P1 using (*)

CPU time₁ = 10⁹ x 0.9 / 4 Ghz

= 0.9 x 10⁹ / 4 x 10⁹ Hz = 0.225 s

So, we need to find the number of instructions₂ such that CPU time₂ = 0.225 s. This means that using (*) along with clock rate₂ = 3 Ghz and CPI₂ = 0.75

Therefore, numbers of instruction₂ x 0.75 / 3 Ghz = 0.225 s

Hence, numbers of instructions₂ = 0.225 x 3 x 10⁹ / 0.75 = 9 x 10⁸

So, P1 can process more instructions than P2 in the same period of time.

(3)

We recall that:

MIPS = Clock rate / CPI X 10⁶

So, MIPS₁ = 4GHZ / 0.9 X 10⁶ = 4 X 10⁹HZ / 0.9 X 10⁶ = 4444

MIPS₂ = 3GHZ / 0.75 X 10⁶ = 3 x 10⁹ / 0.75 X 10⁶ = 4000

So, P1 has the bigger MIPS

(4)

We now recall that:

MFLOPS = FLOPS Instructions / time x 10⁶

= 0.4 x instructions / time x 10⁶ = 0.4 MIPS

Therefore,

MFLOPS₁ = 1777.6

MFLOPS₂ = 1600

Again, P1 has the bigger MFLOPS

3 0

2 years ago