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omeli [17]
3 years ago
11

All data process by a computer must be in 1. binary form 2. Hexadecimal form 3. Duodecimal form 4. Unitary form?

Computers and Technology
2 answers:
Sergeu [11.5K]3 years ago
4 0
Binary form
Those are the crazy 1s and 0s ie (1000100111100)
avanturin [10]3 years ago
3 0

Binary code I believe.

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The answer is Every six months

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_____ is a problem-solving technique where each problem in a database is stored with a description and keywords that identify it
babymother [125]

Answer:

Case-based reasoning.

Explanation:

A database management system (DBMS) can be defined as a collection of software applications that typically enables computer users to create, store, modify, retrieve and manage data or informations in a database. Generally, it allows computer users to efficiently retrieve and manage their data with an appropriate level of security.

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7 0
3 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
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Explanation:

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