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3 years ago

8

The study of how to design software, solve problems such as computer security threats, or come up with better ways of handling d

ata storage

Computers and Technology

1 answer:

Gnom [1K]3 years ago

3 0

Computer engineering

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Peter has recently bought a media player and a digital camera he wants to buy a memory card and then use devices which memory do

Helen [10]

Generally B, most modern consumer cameras use either CF (Compact Flash) or SD cards.

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3 years ago

Read 2 more answers

A nonpipelined system takes 300ns to process a task. The same task can be processed in a 5-segment pipeline with a clock cycle o

Vladimir79 [104]

Answer:

Pipelined architecture is 1.667 times more speedy than non pipelined architecture.

Explanation:

There are five stages to to complete an instruction to execute in pipeline architecture. as

Stage 1: Instruction Fetch

Stage 2: Instruction Decode

Stage 3: Execute

Stage 4: Memory Access

Stage 5 : Write Back

The pipeline architecture processed instruction as given below. Each Stage take 1 clock cycle, which is represented as CC.

Stage 1 Stage 2 Stage 3 Stage 4 Stage 5

I1: 1cc 2cc 3cc 4cc 5cc

I2 2cc 3cc 4cc 5cc 6cc

I3 3cc 4cc 5cc 6cc 7cc

I4 4cc 5cc 6cc 7cc 8cc

each four Instructions set takes 8 Clock Cycles.

I5 9cc 10cc 11cc 12cc 13cc

I6 10cc 11cc 12cc 13cc 14cc

I7 11cc 12cc 13cc 14cc 15cc

I8 12cc 13cc 14cc 15cc 16cc

.

.

.

I97: 193cc 194cc 195cc 196cc 197cc

I98 194cc 195cc 196cc 197cc 198cc

I99 195cc 196cc 197cc 198cc 199cc

I100 196cc 197cc 198cc 199cc 200cc

<u>Pipelined architecture Time Calculation</u>

So in pipeline architecture 100 instructions takes 200 clock cycles to execute.

1 Clock Cycle time period is = 60ns.

200 Clock Cycle = 60ns * 200 = 12000 ns

<u>Non Pipelined architecture Time Calculation</u>

1 task Time period= 300ns

100 task time period = 100 * 300 = 30000 ns

The difference between pipelined and non pipelined architecture time period is = 30000 - 12000 = 18000 ns

<u>Speedup ratio</u>

<em>Speed up ratio= 30000/18000 = 1.667</em>

7 0

3 years ago

What method of the Math object can be used to return the largest value from the values that are passed to it?

miss Akunina [59]

Answer:

the Math.max() method

Explanation:

The java.lang.* package contains all the essential components needed for a JAVA program which involves numbers and all the operations done on the numericals.

The java.lang.math class contains all the methods needed to perform mathematical operations on numerical values.

All the methods in the question are discussed below with examples.

Math.max() method

java.lang.Math.max(double/ int/ float a, double/ int/ float b)

This method returns the highest value from among the parameters passed to it.

The syntax for the above is

Numbers can be integers, float and double or long. The return type is the same as the parameters.

Two values to be compared are passed as arguments and the higher value of them is returned.

This function also accepts negative integers as parameters.

Math.max(-9, -2);

Gives -2 as the output.

Math.max(0.34, 0.45)

Gives 0.45 as the output.

Math.ceil() method

java.lang.Math.ceil(double/ int/ float a)

This method returns the nearest greater whole number for the given number.

Integer parameter returns an integer number.

Math.ceil(7);

Gives 7 as the output.

Decimal number will return the nearest higher integer.

Math.ceil(8.7);

Gives 9.0 as the output.

Math.ceil(-0.34)

Gives -0 as the output.

Math.pow() method

java.lang.Math.pow(double/ float/ int a, double/ float/ int b)

The above method returns the result of a raise to the power of b.

The first parameter is the base while the second parameter forms the exponent.

Math.pow( 2.3, 4.5 );

This calculates (2.3)^4.5.

Gives 42.44‬ as the output.

Math.pow( 2, 3 );

This calculates the value of 2^3.

Gives 8 as the output.

Math.abs() method

java.lang.Math.abs(double a)

This method returns non-negative value of the numerical parameter passed to it. Absolute of a number does not changes its value but changes the sign of a negative number. It returns the positive or negation of the negative input number.

Math.abs(-0.1234)

Gives 0.1234 as the output.

Math.abs(12.34)

Gives 12.34 as the output.

These methods require java.lang.Math to be imported in the program.

7 0

3 years ago

You are a software engineer at a company where management routinely encourages you and your colleagues to use pirated software.

s2008m [1.1K]

Answer:

That’s highly illegal. You should notify them that it is illegal. And, if they continue, you should notify the legal team. If it continues, you either need to leave the company or notify the proper authorities.

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2 years ago

An athlete runs every day for five days. Write a program that computes the total distance and average distance ran by the athlet

Tamiku [17]

Answer:

// here is Hw1_q1_code.c

#include <stdio.h>

// main function

int main(void) {

// variables

float dis_mon,dis_tue,dis_wed,dis_thu,dis_fri;

printf("enter distance ran by athlete on monday:");

// read distance on monday

scanf("%f",&dis_mon);

printf("enter distance ran by athlete on tuesday:");

// read distance on tuesday

scanf("%f",&dis_tue);

printf("enter distance ran by athlete on wednesday:");

// read distance on wednesday

scanf("%f",&dis_wed);

printf("enter distance ran by athlete on thursday:");

// read distance on thursday

scanf("%f",&dis_thu);

printf("enter distance ran by athlete on friday:");

// read distance on friday

scanf("%f",&dis_fri);

// total distance

float sum=dis_mon+dis_tue+dis_wed+dis_thu+dis_fri;

// average distance

float average=sum/5;

// print the total and average

printf("total distance ran by athlete is: %f miles",sum);

printf("\naverage distance ran each day is: %f miles",average);

return 0;

}

Explanation:

Declare five variables to store distance ran by athlete on each day from monday to friday.Read the five distance.Then calculate their sum and assign to variable "sum".Find the average distance ran by athlete by dividing sum with 5 and assign to variable "average".Then print the total distance ran and average distance on each day.

Output:

enter distance ran by athlete on monday:10

enter distance ran by athlete on tuesday:11

enter distance ran by athlete on wednesday:8

enter distance ran by athlete on thursday:9

enter distance ran by athlete on friday:12

total distance ran by athlete is: 50.000000 miles

average distance ran each day is: 10.000000 miles

5 0

3 years ago