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lorasvet [3.4K]
4 years ago
7

Widget Corp. wants to shift its list of inventory to a cloud so that its different branches can access it easily. The company ne

eds a cloud computing option that would provide high flexibility to add or drop resources. The cloud computing option should be cost-effective and should not expose mission-critical applications and data to the outside world. Which cloud computing option would be most suitable for Widget Corp.?
Computers and Technology
1 answer:
____ [38]4 years ago
5 0

Answer:

The best cloud option for Widget Corp considering the cloud computing option should be cost-effective and should not expose mission-critical applications and data to the outside world.

Is a hybrid could.

Explanation:

The reasons behind this answer are that in the first place the cloud is not going to be fully managed by a third party, but also by the IT department of Widget Corp allowing it to control its security. Also, that the cloud can be adapted to public or mainstream sources, resources, and platforms. Making it very user friendly and lowering down specific knowledge to use it.

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A bakery collects data on sales. Each sales record includes the date of the sale and some metadata about the items that were par
professor190 [17]

Answer:

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Explanation:

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The date on which the items are sold

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3 years ago
5.2
Ray Of Light [21]

Answer:

In order to reduce the risk of accidents on the roads.

Explanation:

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Write a function which the counts the number of odd numbers and even numbers currently in the stack and prints the results.
Lerok [7]

Answer:

See the code below and the algorithm explanation on the figure.

Explanation:

The explanation in order to get the answer is given on the figure below.

Solving this problem with C. The program is given below:

#include <stdio.h>

int main(void) {

   int n, Even=0, Odd=0, Zeros=0;  

   for (;;) {

       printf("\nEnter the value the value that you want to check(remember just integers): ");

       //IF we input a non-numeric character the code end;

       if (scanf("%d", &n) != 1) break;

       if (n == 0) {

           Zeros++;

       }

       else {

           if (n % 2) {

               Even++;

           }

           else {

               Odd++;

           }

       }

   }  

   printf("for this case we have %d even, %d odd, and %d zero values.", Even, Odd, Zeros);

   return 0;

}

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