Why are computers useful for modeling situations?

TEA [102]

Answer: A flowchart is a graphical representation of decisions and their results mapped out in individual shapes.

Explanation:

These shapes were first developed by Herman Goldstine and John von Neumann in the 1940s.

Flowcharts can provide a step-by-step diagram for mapping out complex situations, such as programming code or troubleshooting problems with a computer.

4 0

3 years ago

Enumeration can be used to discover all but which of the following types of information?

Ostrovityanka [42]

Answer:A)Smart-card PINs

Explanation: A pin is the defines as the personal identification number that is allotted to the smart card uniquely that is no other smart can have same number.This number is only known by the smart card holder only.

This number is important because it works for the unlocking of the data that is securely stored and by entering the pin ,it can be decoded.Thus, the other given options are incorrect because they don't use enumeration to hide the data .Thus the correct option is option (A)

3 0

3 years ago

What savings account has the best rate of return on his interest

NeX [460]

Ally.com is a great savings account.

8 0

3 years ago

In the following four questions, we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the pa

pantera1 [17]

Answer:

1)we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Initially suppose there is only one link between source and destination. Also suppose that the entire MP3 file is sent as one packet. The TRANSMISSION DELAY is:

3 Seconds

2)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. The END TO END DELAY(transmission delay plus propagation delay) is

3.05 seconds

3)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. how many bits will the source have transmitted when the first bit arrives at the destination.

500,000 bits

4)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there are two links between source and destination, with one router connecting the two links. Each link is 5,000 km long. Again suppose the MP3 file is sent as one packet. Suppose there is no congestion, so that the packet is transmitted onto the second link as soon as the router receives the entire packet. The end-to-end delay is

6.1 seconds

5)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose that the MP3 file is broken into 3 packets, each of 10 Mbits. Ignore headers that may be added to these packets. Also ignore router processing delays. Assuming store and forward packet switching at the router, the total delay is

4.05 seconds

6)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there is only one link between source and destination, and there are 10 TDM channels in the link. The MP3 file is sent over one of the channels. The end-to-end delay is

30.05 seconds

7)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there is only one link between source and destination, and there are 10 FDM channels in the link. The MP3 file is sent over one of the channels. The end-to-end delay is

30.05 seconds

6 0

3 years ago

I'm stuck on this Java exercise problem and don't know how to organize my code with proper formatting and it's not helping that

Volgvan

In your "count spaces method" before the for loop, you want to declare an int variable (let's call it spc)

then, each time you find that charat, spc++.

also, the string must be declared in the main method, not the spaces count method.

The actual code:

public class CountSpaces{

public static void main(String[] args){

String instring = "(your quote here)";

int spaces = calculateSpaces(instring);

System.out.println("The number of spaces is " + spaces);

}

public static int calculateSpaces(String in){

int spc = 0;

for(int i= 0; i<in.length; i++){

if(in.charAt(i)== ' '){

spc++;

}

return spc;

}

that should be it. If you have any questions, feel free to reach out.

4 0

3 years ago