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son4ous [18]
2 years ago
8

Which is a key component to participating in a school-to-work program?

Computers and Technology
1 answer:
Solnce55 [7]2 years ago
4 0

The key component to participating in a school-to-work program is known as Job shadows.

<h3>What are job shadows?</h3>

Job shadowing is known to be a kind of on-the-job training that gives room for an interested employee to follow closely an see or analyze another employee carrying out the work.

Therefore, one can say that the key component to participating in a school-to-work program is known as Job shadows.

Learn more about School from

brainly.com/question/2474525

#SPJ1

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What is the purpose of the ISOWEEKNUM function? determines how many workdays are in a certain week determines how many workdays
ivanzaharov [21]

Answer:

It's D

Explanation:

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50 POINTS How do you express yourself and your creativity through computer science?
Igoryamba

Answer:

1. I express myself through the designing, development and analysis of software and hardware used to solve problems in a variety of business, scientific and social contexts.

2. support the development of general reasoning, problem-solving and communication skills.

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Write the Flowchart to find Even number between 1 to 50<br>​
ASHA 777 [7]

Answer:

See attachment for flowchart

Explanation:

Required

Flowchart to fine even from 1 to 50

The flowchart has been attached.

The rough algorithm (explanation) of the flowchart is as follows.

1. Start

2. Initialize num to 1

3. Check if num is less than or equal to 50

  3.1 If yes

      3.1.1 Check if num is even

      3.1.1.1 If yes

         3.1.1.2 Print num

  3.1.3 Increase num by 1

 3.2 If num is greater than 50

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4. Goto 3

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3 years ago
La sentencia de ASIGNAR es una sentencia de entrada de datos, verdadero o falso ?
Akimi4 [234]

Answer:

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7 0
3 years ago
Suppose you have two arrays of ints, arr1 and arr2, each containing ints that are sorted in ascending order. Write a static meth
telo118 [61]
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.

public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}


So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.

A quick explanation:

We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.

The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.


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3 years ago
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