What would be the output of system.out.println("fun\tny");

Match each vocabulary word to its definition.

Nana76 [90]

Answer:

Column matching given in explanation

1. Intranet : a network of computer within an organization

2. Placeholder: an empty area that reserves space for new content

3. Presentation: information delivered to an audience

4: Presentation Technology: a software application that helps organize and convey information

5. Keynote: the main speech delivered to all in attendance

Explanation:

Intranet:

A small network that exist in the small organization or company to connect all the computers in organization.

Placeholder:

In computer programming, few variables has been defined in the start of program. These variables have no value initially but can be utilized at the time of need. These variable holds some space in memory that is called Placeholder.

Presentation:

A piece of information relevant to some particular topic, which will be delivered to a particular audience is called presentation.

Presentation Technology:

The tools that are used to assist presentation such as projector, slides, boards etc. are called presentation technology.

Keynote:

The main idea of speech that presenter want to deliver to the audience is called Keynote.

4 0

3 years ago

Write a program that reads a floating-point number from the user and prints "zero", "positive" or "negative"

Lorico [155]

```
#!/usr/local/bin/python3

foo = float( input( "Enter a number: " ) )
if( foo < 0.0 ):
 print( "negative" )
elif( foo > 0.0 ):
 print( "positive" )
else
 print( "zero" )

exit( 0 )
```

3 0

3 years ago

Using Word, Maureen is writing an outline of a presentation she plans to give to her company. She will be showing a video during

svlad2 [7]

A copy and paste would be fastest and easiest. As you can pinpoint the exact location to where you want the picture to be.

5 0

3 years ago

Read 2 more answers

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

3 years ago

Differentiate hardware andsoftware interrupts?

Tom [10]

Answer:

-Hardware interrupts are the interrupts that occur due to the external devices.

-Software interrupt arises due to the executing program.

Explanation:

Hardware Interrupts are the raised due to external hardware devices whereas the software interrupts are raised by the executing instruction.
Asynchronized events occur in the hardware interrupts whereas software interrupt face the synchronized events.
There is increase in the program counter in software interrupt but no increment in the program counter for the hardware interrupt.

8 0

3 years ago