Answer:
I would say Henry Ford's invention; the Ford car company, since we still have Ford cars running nowadays.
Explanation:
d. Refers to the depth and breadth of details contained in a piece of textual, graphic, audio, or video information.
<h3>Information richness</h3>
Information richness refers to the amount of sensory input available during communication. For example, <u>talking to co-workers in a monotone voice without varying pace or gestures is not a very enriching experience</u>. On the other hand, using gestures, tone of voice, and pace of speech to convey meaning beyond the words themselves promotes richer communication. Different channels have different information wealth.
Information-rich channels convey more non-verbal information. For example, a face-to-face conversation is richer than a phone call, but a phone call is richer than an email. Research shows that effective managers are more likely to use informative channels of communication than ineffective managers.
Learn more about Note-making Information: brainly.com/question/1299137
#SPJ4
Answer:
Focus Stacking
Explanation:
F-stop stacking which is also referred to as image stacking is a powerful technique that improves the quality of an image by stacking images taken at different f-stops in order to improve corner sharpness and overcome blurriness. Once photographs are taken at different f-stops (focus stops), a final complete composite picture is created using only the sharpest portions of the photograph.
Answer:
Explanation:
The following code is written in Java and is a function/method that takes in an int array as a parameter. The type of array can be changed. The function then creates a counter and loops through each element in the array comparing each one, whenever one element is found to be a duplicate it increases the counter by 1 and moves on to the next element in the array. Finally, it prints out the number of duplicates.
public static int countDuplicate (int[] arr) {
int count = 0;
for(int i = 0; i < arr.length; i++) {
for(int j = i + 1; j < arr.length; j++) {
if(arr[i] == arr[j])
count++;
}
}
return count;
}
❤️The answer is supply chain effectiveness.
Hope you have a great day!
