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3 years ago

11

People are able to predict future events to some extent because: (Select all that apply)

Computers and Technology

1 answer:

WINSTONCH [101]3 years ago

8 0

According to sources, the most probable answer to this query is

people understand some of the physical laws of nature

all his or her past predictions are true
Thank you for your question. Please don't hesitate to ask in Brainly your queries.

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What is output? public class MathRecursive { public static void myMathFunction(int a, int r, int counter) { int val; val = a*r;

ycow [4]

Answer:

The output of the program is:

2 4 8 16 32 64 End

Explanation:

See attachment for proper presentation of the program

The program uses recursion to determine its operations and outputs.

The function is defined as: myMathFunction(int a, int r, int counter)

It initially gets the following as its input from the main method

a= 1; r = 2; counter = 0

Its operation goes thus:

val = a*r; 1 * 2 = 2

Print val; Prints 2

if (counter > 4) { System.out.print("End"); } : Not true

else { myMathFunction(val, r, counter + 1); }: True

The value of counter is incremented by 1 and the function gets the following values:

a= 2; r = 2; counter = 1

val = a*r; 2 * 2 = 4

Print val; Prints 4

else { myMathFunction(val, r, counter + 1); }: True

The value of counter is incremented by 1 and the function gets the following values:

a= 4; r = 2; counter = 2

val = a*r; 4 * 2 = 8

Print val; Prints 8

else { myMathFunction(val, r, counter + 1); }: True

The value of counter is incremented by 1 and the function gets the following values:

a= 8; r = 2; counter = 3

val = a*r; 8 * 2 = 16

Print val; Prints 16

else { myMathFunction(val, r, counter + 1); }: True

The value of counter is incremented by 1 and the function gets the following values:

a= 16; r = 2; counter = 4

val = a*r; 16 * 2 = 32

Print val; Prints 32

else { myMathFunction(val, r, counter + 1); }: True

The value of counter is incremented by 1 and the function gets the following values:

a= 32; r = 2; counter = 5

val = a*r; 32 * 2 = 64

Print val; Prints 64

if (counter > 4) { System.out.print("End"); } : True

<em>This prints "End"</em>

So; the output of the program is:

2 4 8 16 32 64 End

3 0

2 years ago

The _______ provides access to the internet; may also be internal?

VikaD [51]

World Wide Web (WWW)? I honestly don't know.

8 0

3 years ago

Задача в том, чтобы считать

Levart [38]

I’m not sure what you are saying

5 0

3 years ago

What’s the difference between the alternative press and a tabloid?

miskamm [114]

Answer:

Alternative Press just differs in opinion from mainstream media. A tabloid is literally smaller and is contains sensationalist and largely photographic content.

Most alternative presses aren't reliable and are effectively tabloids in nature, but their content differs.

4 0

2 years ago

We begin with a computer implemented in single-cycle implementation. When the stages are split by functionality, the stages do n

vampirchik [111]

<u>Answer:</u>

a) First, we need to determine the pipeline stage amounting to the maximum time. In the given case, the maximum time required is 2ns for MEM. In addition, the pipeline register delay=0.1 ns.

Clock cycled time of the pipelined machine= max time+delay

=2ns+0.1 ns

=2.1 ns

b) For any processor, ideal CPI=1. However, since there is a stall after every four instructions, the effective CPI of the new machine is specified by:

$1+(1 / 4)=1.25$

c) The speedup of pipelined machine over the single-cycle machine=avg time per instruction of single cycle/avg time per instruction of pipelined.

Single cycle processor:

CPI=1

Clock period=7 ns

Pipelined processor:

Clock period=2.1 ns

CPI=1.25

Therefore, speedup= $=7^{*} 1 /\left(2.1^{*} 1.25\right)$

=7/2.625

= 2.67

d) As the number of stages approach infinity, the speedup=k where k is the number of stages in the machine.

8 0

3 years ago