A tasks are suitable for creating an algorithm are:
- giving directions to a location.
- solving a math problem.
- tracking money in a bank account.
- tracking the number of items in inventory.
<h3>What is algorithm?</h3>
An algorithm is known to be a form of a procedure that is often employed in the act of solving a problem or carrying out a computation.
Note that in the case above, A tasks are suitable for creating an algorithm are:
- giving directions to a location.
- solving a math problem.
- tracking money in a bank account.
- tracking the number of items in inventory.
See options below
giving directions to a location
saving time writing a computer program
solving a math problem
tracking money in a bank account
tracking the number of items in inventory
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Answer:
print("hello world")
Explanation:
a hello world program is simply a program that prints out hello world.
for this you would need to remember to have the same number of brackets on each side and to write print. Also remember when printing to include speech marks.
Answer:
Explanation:
Keep in mind a lossy algorithm will lose information while a lossless algorithm maintains all your original information.
Therefore:
A. False, a lossy algorithm will not allow perfect reconstruction.
B. True, if you don't care about keeping all your information it's easier to compress.
C. False, you can use a lossless algorithm for anything.
D. False, the point of lossless is that you keep all information.
answer:
Mainframe computers are used by large companies and organisations to perform critical tasks that involve bulk data processing like transaction processing, census information, statistical data and so on. They consist of extensive input and output facilities, are very stable and dependable and handle millions of transactions every day
Explanation:
left[0]=a[0];
for(int i=1;i<=n-1;i++)
left[i]=(left[i-1]*a[i])%M;
right[n-1]=a[n-1];
for(int i=n-2;i>=0;i--)
right[i]=(right[i-1]*a[i])%M;
for query q
if(q==0)
return right[1]%M;
if(q==n-1)
return left[n-2]%M;
return (left[q-1]*right[q+1])%M;
