Lesser-known ideas and interpretations of common knowledge need to be documented. true or false.

Which of the selections below does not represent a workable ip address?

defon

Answer: C

Explanation:

Because ipv4 ip addresses will be 4 sets of numbers for a maximum of 255 per set ie. The highest ipv4 number is 255.255.255.255 and the answer C has 293 in it.

6 0

4 years ago

Escribe todas las posibles combinaciones que pueden haber entre 4 bits.

Vedmedyk [2.9K]

Answer:

The answer is " $\bold{2^n\ \ _{where} \ \ \ n \ is\ bit }$ "

Explanation:

The Combinations could be produced by using n-bits $2^n$ . It s enables you to generate the 4-bit numbers that are:

$\to 2^4 = 16$ combinations.

for the 4-bit, the combination of 2 = 16, which are its possible combination and for the 10 variations appropriate 16 combinations are used, As we know for 4 bit 16 Combinations can be generated which are from 0 to 15.

7 0

3 years ago

A person gets 13 cards of a deck. Let us call for simplicity the types of cards by 1,2,3,4. In How many ways can we choose 13 ca

natima [27]

Answer:

There are 5,598,527,220 ways to choose 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 cards of type 4 from a set of 13 cards.

Explanation:

The crucial point of this problem is to understand the possible ways of choosing any type of card from the 13-card deck.

This is a problem of combination since the order of choosing them does not matter here, that is, the important fact is the number of cards of type 1, 2, 3 or 4 we can get, no matter the order that they appear after choosing them.

So, the question for each type of card that we need to answer here is, how many ways are there of choosing 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 are of type 4 from the deck of 13 cards?

The mathematical formula for combinations is $\\ \frac{n!}{(n-k)!k!}$ , where n is the total of elements available and k is the size of a selection of k elements from which we can choose from the total n.

Then,

Choosing 5 cards of type 1 from a 13-card deck:

$\frac{n!}{(n-k)!k!} = \frac{13!}{(13-5)!5!} = \frac{13*12*11*10*9*8!}{8!*5!} = \frac{13*12*11*10*9}{5*4*3*2*1} = 1,287$ , since $\\ \frac{8!}{8!} = 1$ .

Choosing 4 cards of type 2 from a 13-card deck:

$\\ \frac{n!}{(n-k)!k!} = \frac{13!}{(13-4)!4!} = \frac{13*12*11*10*9!}{9!4!} = \frac{13*12*11*10}{4!}= 715$ , since $\\ \frac{9!}{9!} = 1$ .

Choosing 2 cards of type 3 from a 13-card deck:

$\\ \frac{n!}{(n-k)!k!} =\frac{13!}{(13-2)!2!} = \frac{13*12*11!}{11!2!} = \frac{13*12}{2!} = 78$ , since $\\ \frac{11!}{11!}=1$ .

Choosing 2 cards of type 4 from a 13-card deck:

It is the same answer of the previous result, since

$\\ \frac{n!}{(n-k)!k!} = \frac{13!}{(13-2)!2!} = 78$ .

We still need to make use of the Multiplication Principle to get the final result, that is, the ways of having 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 cards of type 4 is the multiplication of each case already obtained.

So, the answer about how many ways can we choose 13 cards so that there are 5 of type 1, there are 4 of type 2, there are 2 of type 3 and there are 2 of type 4 is:

1287 * 715 * 78 * 78 = 5,598,527,220 ways of doing that (or almost 6 thousand million ways).

In other words, there are 1287 ways of choosing 5 cards of type 1 from a set of 13 cards, 715 ways of choosing 4 cards of type 2 from a set of 13 cards and 78 ways of choosing 2 cards of type 3 and 2 cards of type 4, respectively, but having all these events at once is the multiplication of all them.

5 0

4 years ago

All who are interested in forex trading and bitcoin mining follow me for account management to all who are busy with work so i c

masya89 [10]

I am very interested but too young just 10

8 0

3 years ago

Refer to the method f:

Goshia [24]

<h2>This function will land up in infinite function call</h2>

Explanation:

first time when the function gets invoked,

f(6,8), so k=6 & n=8, inside the function it checks k==n, ie. 6==8, returns false, then one more if is available, so 6>8 is check for , once again it is false and else loop is executed, the function is called recursively using f(k-n,n), that is f(6-8,8), it means f(-2,8) is passed.

Second time,

if(-2==8) is false, so if(-2>8) is again false and function f(-10, 8) is called

if(-10==8) is false, so if(-10>8) is again false and function f(-18,8) is called

if(-18==8) is false, so if(-18>8) is again false and function f(-26,8) is called

So this goes recursively and ends in an infinite function call.

5 0

4 years ago