Answer:
The code is given below
Explanation:
The correct syntax would be to place appropriate parenthesis.
(month==1?"jan":(month==2?"feb":(month==3?"mar":(month==4?"apr":(month==5?"may":(month==6?"jun":(month==7?"jul":(month==8?"aug":(month==9?"sep":(month==10?"oct":(month==11?"nov":"dec")))))))))));
Similarly, you can also use the following code:
String[] months = { "jan", "feb", "mar", "apr", "may", "jun", "jul", "aug", "sep", "oct", "nov", "dec" };
int month = 1;
String monthDescription = months[month - 1];
Answer:
Pseudocode is as follows:
// below is a function that takes two parameters:1. An array of items 2. An integer for weight W
// it returns an array of selected items which satisfy the given condition of sum <= max sum.
function findSubset( array items[], integer W)
{
initialize:
maxSum = 0;
ansArray = [];
// take each "item" from array to create all possible combinations of arrays by comparing with "W" and // "maxSum"
start the loop:
// include item in the ansArray[]
ansArray.push(item);
// remove the item from the items[]
items.pop(item);
ansArray.push(item1);
start the while loop(sum(ansArray[]) <= W):
// exclude the element already included and start including till
if (sum(ansArray[]) > maxSum)
// if true then include item in ansArray[]
ansArray.push(item);
// update the maxSum
maxSum = sum(ansArray[items]);
else
// move to next element
continue;
end the loop;
// again make the item[] same by pushing the popped element
items.push(item);
end the loop;
return the ansArray[]
}
Explanation:
You can find example to implement the algorithm.
This is the recorded history of the development of a group functioning within an organization. Identify the sentence that refers to the norming stage of group development.
. The group then created a working structure that everyone agreed with.
Answer:
Let's convert the decimals into signed 8-bit binary numbers.
As we need to find the 8-bit magnitude, so write the powers at each bit.
<u>Sign -bit</u> <u>64</u> <u>32</u> <u>16</u> <u>8</u> <u>4</u> <u>2</u> <u>1</u>
+25 - 0 0 0 1 1 0 0 1
+120- 0 1 1 1 1 0 0 0
+82 - 0 1 0 1 0 0 1 0
-42 - 1 0 1 0 1 0 1 0
-111 - 1 1 1 0 1 1 1 1
One’s Complements:
+25 (00011001) – 11100110
+120(01111000) - 10000111
+82(01010010) - 10101101
-42(10101010) - 01010101
-111(11101111)- 00010000
Two’s Complements:
+25 (00011001) – 11100110+1 = 11100111
+120(01111000) – 10000111+1 = 10001000
+82(01010010) – 10101101+1= 10101110
-42(10101010) – 01010101+1= 01010110
-111(11101111)- 00010000+1= 00010001
Explanation:
To find the 8-bit signed magnitude follow this process:
For +120
- put 0 at Sign-bit as there is plus sign before 120.
- Put 1 at the largest power of 2 near to 120 and less than 120, so put 1 at 64.
- Subtract 64 from 120, i.e. 120-64 = 56.
- Then put 1 at 32, as it is the nearest power of 2 of 56. Then 56-32=24.
- Then put 1 at 16 and 24-16 = 8.
- Now put 1 at 8. 8-8 = 0, so put 0 at all rest places.
To find one’s complement of a number 00011001, find 11111111 – 00011001 or put 0 in place each 1 and 1 in place of each 0., i.e., 11100110.
Now to find Two’s complement of a number, just do binary addition of the number with 1.
She would use Ctrl and C bc thats the copy short cut
Ctrl and v is pasting
Ctrl and a is selecting all text and
Ctrl and z is for cutting the text
