List 2 end to end test commands. <br><br> Will mark Brainliest!!

What is the key to satisfaction and success when choosing a career

ELEN [110]

Answer:

be yourself

Explanation:

3 0

3 years ago

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Find the propagation delay for a signal traversing the in a metropolitan area through 200 km, at the speed of light in cable (2.

Ede4ka [16]

Answer:

t= 8.7*10⁻⁴ sec.

Explanation:

If the signal were able to traverse this distance at an infinite speed, the propagation delay would be zero.

As this is not possible, (the maximum speed of interactions in the universe is equal to the speed of light), there will be a finite propagation delay.

Assuming that the signal propagates at a constant speed, which is equal to 2.3*10⁸ m/s (due to the characteristics of the cable, it is not the same as if it were propagating in vaccum, at 3.0*10⁸ m/s), the time taken to the signal to traverse the 200 km, which is equal to the propagation delay, can be found applying the average velocity definition:

$v = \frac{(xf-xo)}{(t-to)}$

If we choose x₀ = 0 and t₀ =0, and replace v= 2.3*10⁸ m/s, and xf=2*10⁵ m, we can solve for t:

$t =\frac{xf}{v} =\frac{2e5 m}{2.3e8 m/s} =8.7e-4 sec.$

⇒ t = 8.7*10⁻⁴ sec.

4 0

3 years ago

Create an array to hold the rainfall values. Create a 2nd parallel array (as a constant) to hold the abbreviated names of the mo

Zarrin [17]

Answer:

#include <stdio.h>

int main()

{

//variable declaration

int low, high;

float lowRain, highRain, total, avg;

//array declaration

float rainfall[13];

char monthName[13][10] = {"", "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"};

//get user input

for(int i=1; i<=12; i++)

{

printf("Enter the rainfall (in inches) for %s: ", monthName[i]);

scanf("%f", &rainfall[i]);

}

//display the monthly rainfall

printf("\nThe rainfall that was entered was:\n");

for(int i = 1; i<=6; i++)

printf("%s ", monthName[i]);

printf("\n");

for(int i = 1; i<=6; i++)

printf("%.1f ", rainfall[i]);

printf("\n");

for(int i = 7; i<=12; i++)

printf("%s ", monthName[i]);

printf("\n");

for(int i = 7; i<=12; i++)

printf("%.1f ", rainfall[i]);

//variable initialization

low = 1;

high = 1;

lowRain = rainfall[1];

highRain = rainfall[1];

total = 0;

//calculate the lowest, highest and averaage rainfall

for(int i=1; i<=12; i++)

{

if(lowRain>rainfall[i])

{

lowRain = rainfall[i];

low = i;

}

if(highRain<rainfall[i])

{

highRain = rainfall[i];

high = i;

}

total = total + rainfall[i];

}

avg = total / 12;

//display the result

printf("\n\nThe total rain that fell was %.1f inches", total);

printf("\nThe average monthly rainfall was %.1f inches.", avg);

printf("\nThe lowest monthly rainfall was %.1f inches in %s.", rainfall[low], monthName[low]);

printf("\nThe highest monthly rainfall was %.1f inches in %s.", rainfall[high], monthName[high]);

return 0;

}

4 0

3 years ago

What is the importance of the international employment

My name is Ann [436]

Answer:

People who went overseas to work can learn different skills and technologies which can be beneficial for the development of our own country.

7 0

3 years ago

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Consider the following method: public double doubleVal(double x) { return x *2; } The following code segment calls the method do

VashaNatasha [74]

Answer:

13.0

Explanation:

The method doubleVal() is created to accept a single parameter of type double.

It multiplies what ever the value of the parameter is and returns the resulting value.

In this question The method is called within this output statement System.out.println(doubleVal(val)); (Note that val had already been declared and assigned the value of 6.5)

The value 6.5 is doubled and outputed to the screen

8 0

3 years ago

List 2 end to end test commands. Will mark Brainliest!!