Cisco Packet Tracer is tool used for network simulation and visualization program.
The two methods can be used to access and modify an existing program that is running on an IoT device in Cisco packet trace are the following:
1. Click on the device then select the Programming tab.
2. Go to the registration server and login. Then select the Editor tab.
I've included my code in the picture below. Best of luck
Answer:
n! = n*(n-1)*(n-2)*(n-3)* ... *2*1
Explanation:
The factorial operator is simply a mathematical expression of the product of a stated integer and all integers below that number down to 1. Consider these following examples:
4! = 4 * 3 * 2 * 1
4! = 12 * 2 * 1
4! = 24
6! = 6 * 5 * 4 * 3 * 2 * 1
6! = 30 * 4 * 3 * 2 * 1
6! = 120 * 3 * 2 * 1
6! = 360 * 2 * 1
6! = 720
So, the factorial of n would follow the same as such:
n! = n * (n-1) * (n-2) * ... * 2 * 1
Cheers.
Answer:
have the same SSID but different channels
Explanation:
Based on the information provided within the question it can be said that the best option to accomplish this would be to have the same SSID but different channels. This would maintain the users connected to the same network name (SSID) but still be able to roam and jump from one access point to the other.
Answer:
d. public myClass( ) {. . .}
Explanation:
A constructor is a special method that is called when an object of a class is created. It is also used to initialize the instance variables of the given class. A class may have one or more constructors provided that these constructors have different signatures. A class that does not have a constructor explicitly defined has a default parameterless constructor.
Having said these about a constructor, a few other things are worth to be noted by a constructor.
i. In Java, a constructor has the same name as the name of its class.
For example, in the given class <em>myClass</em>, the constructor(s) should also have the name <em>myClass</em>.
ii. A constructor does not have a return value. It is therefore wrong to write a constructor like this:
<em>public void myClass(){...}</em>
This makes option a incorrect.
iii. When a constructor with parameters is defined, the default parameterless constructor is overridden. This might break the code if some other parts of the program depend on this constructor. So it is advisable to always explicitly write the default parameterless constructor.
This makes option d a correct option.
Other options b and c may also be correct but there is no additional information in the question to help establish or justify that.
