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Sunny_sXe [5.5K]
3 years ago
5

Which of these is used at testing/development stage?

Computers and Technology
1 answer:
maw [93]3 years ago
3 0

Answer:

Which of these is used at testing/development stage?

A. compiler

B. interpreter

<u>C. JAVA </u>

D. Arduino

Hope this helps :)

Have a great day !

5INGH

Explanation:

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Can a computer will work more efficiently if you perform disk optimization
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List two reasons why “buying a computer is no easy task” ?
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Imagine that a crime has been committed and being a police officer you have to investigate the case. Solving a crime can be a ve
Igoryamba

Explanation:

On October 31, around twelve o'clock at night, the lights were dull. It was Siberian cold and I was walking painfully under the whitish lighting of a lamppost and the retro neon lights of a convenience store in a neighborhood that was buzzing when I went to buy meat. buildings, grayish in this half-light, crumbled under the weight of sleep, while the neighborhood was still open. Its decoration was particularly macabre in tone: synthetic but very sticky spider webs adorned the porches of the terraces, pumpkin ornaments are exhibited like works of art, ... Some pushed the macabre by hanging mannequins made up in corpses. At the entrance to the neighborhood, I found the body of its owner stabbed. Frozen in place, I wondered what else awaits me. My heart was pounding, feeling a shiver of anguish. When I was removing the phone from my pocket to call my police friends, my hands were clenched in fear. An investigation was carried out to find out the causes of death. Also, I went with them so they take my words. Hopefully the murderer will be arrested to put an end to this sybilinous crime.

4 0
3 years ago
Computer A has an overall CPI of 1.3 and can be run at a clock rate of 600 MHz. Computer B has a CPI of 2.5 and can be run at a
kirill115 [55]

Answer:

Check the explanation

Explanation:

CPI means Clock cycle per Instruction

given Clock rate 600 MHz then clock time is Cー 1.67nSec clockrate 600M

Execution time is given by following Formula.

Execution Time(CPU time) = CPI*Instruction Count * clock time = \frac{CPI*Instruction Count}{ClockRate}

a)

for system A CPU time is 1.3 * 100, 000 600 106

= 216.67 micro sec.

b)

for system B CPU time is =\frac{2.5*100,000}{750*10^6}

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c) Since the system B is slower than system A, So the system A executes the given program in less time

Hence take CPU execution time of system B as CPU time of System A.

therefore

216.67 micro = =\frac{2.5*Instruction}{750*10^6}

Instructions = 216.67*750/2.5

= 65001

hence 65001 instruction are needed for executing program By system B. to complete the program as fast as system A

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