Answer:
C. deriving benefits from highly focused and high technology markets
Explanation:
Firms that engage in cost-leadership approach seek to combine a low per-unit income with large sales for profit making purposes. Typically, but not always, they are more inclined to market their products and services to a large population base or a niche with a high demand volume. While differentiation enables a company to accomplish a competitive advantage. A Competitive advantage enables a company to achieve more strides over other companies offering related product substitutes. It is an important marketing process that is of critical economic importance to a business.
It should be noted that deriving benefits from highly focused and high technology markets is not part of the approaches to combining overall cost leadership and differentiation competitive advantages.
Answer:
text labels and different
Explanation:
Answer:
Answer is A network access method that avoids the possibility of data collisions.
Token Ring is the precursor to Ethernet and CSMA/CD to avoid packet collisions.
Explanation:
Answer:
Option C i.e., RIPEMD is the correct option for the following question.
Explanation:
Because RIPEMD is the algorithm that is used for the hash primary design characteristic which has two independent and unlike parallel chains of the calculation and computation and the outputs of which are then linked to the process at the end. It is also used for the safe process of replacing the 128-bit hash method.
Other options are wrong because these are not related to the following statement.
Answer:
import java.util.ArrayList;
import java.util.Collections;
class Bulgarian
{
public static void main(String[] args)
{
max_cards=45;
arr->new ArraryList
col=1;
card=0;
left=max_cards;
do{
col->random number
row->new ArrayList;
for i=0 to i<col
{
card++
add card into row
}
add row into arr
left=max_cards-card;
}while(card<max_cards);
finish->false
while(!finish)
{
for i=0 to arr.size()
remove one card from each row
add row into arr
size->new ArrayList;
for i=0 to i<arr.size()
add arr[i] into size
sort(size)
for i=1 to size.size()
if ith element is equals to (i-1)+1
finish->true
else
finish->false
for i=0 to i<arr.size()
print->arr elements
}
}
}
