One of the best ways to shoot a picture is to frame the subject in the middle of the screen or viewer. T or F

Create a static method that: - is called appendPosSum - returns an ArrayList - takes one parameter: an ArrayList of Integers Thi

Flura [38]

Answer:

The method in Java is as follows:

public static ArrayList<Integer> appendPosSum(ArrayList<Integer> nums) {

int sum = 0;

ArrayList<Integer> newArr = new ArrayList<>();

for(int num : nums) {

if(num>0){

sum+=num;

newArr.add(num); } }

newArr.add(sum);

return newArr;

}

Explanation:

This defines the method; it receives an integer arraylist as its parameter; nums

public static ArrayList<Integer> appendPosSum(ArrayList<Integer> nums) {

This initializes sum to 0

int sum = 0;

This declares a new integer arraylist; newArr

ArrayList<Integer> newArr = new ArrayList<>();

This iterates through nums

for(int num : nums) {

If current element is greater than 0

if(num>0){

This sum is taken

sum+=num;

And the element is added to newArr

newArr.add(num); } }

At the end of the iteration; this adds the calculated sum to newArr

newArr.add(sum);

This returns newArr

return newArr;

}

3 0

3 years ago

Is it true that if the user log out the computer will turn off automatically

sergiy2304 [10]

Yes that is a true statement

8 0

2 years ago

Read 2 more answers

It takes 2 seconds to read or write one block from/to disk and it also takes 1 second of CPU time to merge one block of records.

Alexxx [7]

Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

For optimal 4-way merging, we need one dummy run with size 0.

Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows $L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300$
Merge the run as made in step 1 with the remaining 3 runs bearing length 1000, 1200, 1500. The merged run length is 6000 and is calculated as follows

$L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000$

The resulting run has length 6000 records.

Part b

For step 1

Input Output Time

Input Output Time is given as

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}$

Here

L_run is 2300 for step 01
Size_block is 100 as given
Time_{I/O per block} is 2 sec

So

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec$

So the input/output time is 46 seconds for step 01.

CPU Time

CPU Time is given as

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}$

Here

L_run is 2300 for step 01
Size_block is 100 as given
Time_{CPU per block} is 1 sec

So

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec$

So the CPU time is 23 seconds for step 01.

Total time in step 01

$T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\$

Total time in step 01 is 69 seconds.

For step 2

Input Output Time

Input Output Time is given as

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}$

Here

L_run is 6000 for step 02
Size_block is 100 as given
Time_{I/O per block} is 2 sec

So

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec$

So the input/output time is 120 seconds for step 02.

CPU Time

CPU Time is given as

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}$

Here

L_run is 6000 for step 02
Size_block is 100 as given
Time_{CPU per block} is 1 sec

So

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec$

So the CPU time is 60 seconds for step 02.

Total time in step 02

$T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\$

Total time in step 02 is 180 seconds

Merging Time (Total)

Now the total time for merging is given as

$T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\$

Total time in merging is 249 seconds seconds

5 0

3 years ago

Someone who is young, lacks funds, and really wants to gain technical skills while serving his or her nation should consider

Scilla [17]

Hello there!

They already gave you the answer just by saying "while serving his or her nation". So, the only way you can serve your nation is when you join the military.

So, the correct missing word is Joining the military.

I hope this helps!

6 0

3 years ago

How does making a phone call differ when using: A public phone A cell phone

rosijanka [135]

Answer:

Differ in the quality of their transmission.

Explanation:

For example, when using the public phone box, wired transmission is done not wireless as in cell phones. Public phone box use electronic signals transmitted through a cable network to send voice data which are not very effective for long distance communication.

However, cell phones wirelessly send electromagnetic wave signal to a cell tower close to the caller and then the information is then transmitted to cell tower close to the receiver within a split of a second. This method of communication has much advantages.

7 0

3 years ago