How to set the maximum to 10? Java

Samm1E49 ok my username is mosarider489 also if others want to join u can friend me

mixer [17]

Answer:

ok

Explanation:

3 0

3 years ago

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As Kaydence reads a chapter in her anthropology text, she draws a network diagramming the relationships among the concepts descr

german

Answer:

visual encoding

Explanation:

According to my research on the three types of encoding methods, I can say that based on the information provided within the question Kaydence is best described as capitalizing on visual encoding. Which is the act of associating a certain something with a picture in order to store it into memory, as opposed of associating it with sound or words. In this situation Kaydence is associating it with a networking diagram she drew.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

5 0

3 years ago

Suppose you wanted to run a web server or ftp server from your home. what type of ip address would you want?

Greeley [361]

You would need to have a static ip address.

3 0

3 years ago

Rajesh is considering whether to use a web app to collaborate with a group of friends on a project. What is one reason he should

vivado [14]

Rajesh might want to avoid the use of a web app for the project due to connectivity issues. Hence, one reason he might want to avoid using the Webb app is slow or poor internet connection.

Collaborating on a web application relies very heavily on network connectivity which is the what establishes the link ms allows the apllications to achieve optimum performance.

Therefore, slow internet connection is a major reason whuhe might want to avoid using a web app.

Learn more : brainly.com/question/25531734

6 0

3 years ago

We have said that the average number of comparisons need to find a target value in an n-element list using sequential search is

bija089 [108]

Answer:

Part a: If the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b: If the list contains n elements (where n is even) the middle terms are at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c: The average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

Explanation:

Suppose the list is such that the starting index is 0.

Part a

If list has 15 elements, the middle item would be given at 7th index i.e.

there are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(8,9,10,11,12,13,14) above it. It will have to run 8 comparisons to find the middle term.

If list has 17 elements, the middle item would be given at 8th index i.e.

there are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(9,10,11,12,13,14,15,16) above it.It will have to run 9 comparisons to find the middle term.

If list has 21 elements, the middle item would be given at 10th index i.e.

there are 10 indices (0,1,2,3,4,5,6,7,8,9) below it and 10 indices (11,12,13,14,15,16,17,18,19,20) above it.It will have to run 11 comparisons to find the middle term.

Now this indicates that if the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b

If list has 16 elements, there are two middle terms as one at would be at 7th index and the one at 8th index .There are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(9,10,11,12,13,14,15) above it. It will have to run 9 comparisons to find the middle terms.

If list has 18 elements, there are two middle terms as one at would be at 8th index and the one at 9th index .There are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(10,11,12,13,14,15,16,17) above it. It will have to run 10 comparisons to find the middle terms.

If list has 20 elements, there are two middle terms as one at would be at 9th index and the one at 10th index .There are 9 indices(0,1,2,3,4,5,6,7,8) below it and 9 indices(11,12,13,14,15,16,17,18,19) above it. It will have to run 11 comparisons to find the middle terms.

Now this indicates that if the list contains n elements (where n is even) the middle terms are at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c

So the average number of comparisons is given as

((n+1)/2+(n+2)/2)/2=(2n+3)/4

So the average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

6 0

3 years ago