_____ remove the part of an image starting from an edge

Write the definition of a function power_to, which receives two parameters. The first is a float and the second is an integer. T

Ber [7]

Answer:

public static double power_2(double num1, int num2){
if(num2<0){
return 0;
}
else{
double returnedVal = Math.pow(num1,num2);
return returnedVal;
}
}

Explanation:

This is implemented in Java programming language
Line 1 Is the method declaration which indicates that it will return a floating point number (type double) and accepts two parameters, the first a double and the second an int as required by the question
Line 2 Uses and if statement to check if the second parameter is negative (less than 0) and returns 0.0 if this is true
Lines 5-7 is the else condition that uses java's Math.pow method to raise the num1 to the power num2 and returns the value

3 0

2 years ago

Suppose a host has a 1-MB file that is to be sent to another host. The file takes 1 second of CPU time to compress 50%, or 2 sec

kozerog [31]

Answer: bandwidth = 0.10 MB/s

Explanation:

Given

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.50 MB / Bandwidth)

Transfer Size = 0.50 MB

Total Time = Compression Time + RTT + (0.50 MB /Bandwidth)

Total Time = 1 s + RTT + (0.50 MB / Bandwidth)

Compression Time = 1 sec

Situation B:

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.40 MB / Bandwidth)

Transfer Size = 0.40 MB

Total Time = Compression Time + RTT + (0.40 MB /Bandwidth)

Total Time = 2 s + RTT + (0.40 MB / Bandwidth)

Compression Time = 2 sec

Setting the total times equal:

1 s + RTT + (0.50 MB / Bandwidth) = 2 s + RTT + (0.40 MB /Bandwidth)

As the equation is simplified, the RTT term drops out(which will be discussed later):

1 s + (0.50 MB / Bandwidth) = 2 s + (0.40 MB /Bandwidth)

Like terms are collected:

(0.50 MB / Bandwidth) - (0.40 MB / Bandwidth) = 2 s - 1s

0.10 MB / Bandwidth = 1 s

Algebra is applied:

0.10 MB / 1 s = Bandwidth

Simplify:

0.10 MB/s = Bandwidth

The bandwidth, at which the two total times are equivalent, is 0.10 MB/s, or 800 kbps.

(2) . Assume the RTT for the network connection is 200 ms.

For situtation 1:

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 1 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.50 MB

Total Time = 1.2 sec + 5 sec

Total Time = 6.2 sec

For situation 2:

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 2 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.40 MB

Total Time = 2.2 sec + 4 sec

Total Time = 6.2 sec

Thus, latency is not a factor.

5 0

2 years ago

How many solutions exist for the given equation?

vovikov84 [41]

Infinity amount of answers because each side can be simplified to x making any solution correct

6 0

2 years ago

In the graph shown here, by what percentage are the number of people in computer occupations in general projected to increase?

SOVA2 [1]

Answer:

12% or 21%

Explanation:

not really sure

8 0

2 years ago

Read 2 more answers

Replace any alphabetic character with '_' in 2-character string passCode. Ex: If passCode is "9a", output is:9_Hint: Use two if

Mashcka [7]

Answer:

Following is given the code as required.

I hope it will help you!

Explanation:

7 0

2 years ago

_____ remove the part of an image starting from an edge​

_____ remove the part of an image starting from an edge