Answer:
limited problem solving
Explanation:
Henri will probably use a limited problem-solving process, we reach this conclusion due to two factors:
- The time he has to take the decision is short, therefore he can calmly evaluate his options
- His past positive experience in buying books online will influence his decision
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Answer:
The answer to this question is given below in the explanation section.
Explanation:
WHILE loops and DO WHILE loops are called the condition controlled loops. The execution of these loops depends on a certain condition, when the condition is true, the loop body will be executed and when the condition becomes false, the loop body will not be executed. the major difference between both loops is given below.
In the WHILE loop, the condition is checked at the beginning of the loop whereas in the do-while loop condition is checked at the end of the loop and, in the do-while loop, the loop body is executed at least once.
The syntax of the while loop is given below
while (condition) {
// code block to be executed
}
The syntax of do-while loop is given below
do {
// code block to be executed
}
while (condition);
The nutrition characteristics of eating darker meat of fish
like Salmon are that it is higher in healthy Omega-3 fats. It is true, however,
that as much as it has the highest Omega-3 fats, it is also likely to be
highest in any potential toxins.
Answer:
The page field is 8-bit wide, then the page size is 256 bytes.
Using the subdivision above, the first level page table points to 1024 2nd level page tables, each pointing to 256 3rd page tables, each containing 64 pages. The program's address space consists of 1024 pages, thus we need we need 16 third-level page tables. Therefore we need 16 entries in a 2nd level page table, and one entry in the first level page table. Therefore the size is: 1024 entries for the first table, 256 entries for the 2nd level page table, and 16 3rd level page table containing 64 entries each. Assuming 2 bytes per entry, the space required is 1024 * 2 + 256 * 2 (one second-level paget table) + 16 * 64 * 2 (16 third-level page tables) = 4608 bytes.
First, the stack, data and code segments are at addresses that require having 3 page tables entries active in the first level page table. For 64K, you need 256 pages, or 4 third-level page tables. For 600K, you need 2400 pages, or 38 third-level page tables and for 48K you need 192 pages or 3 third-level page tables. Assuming 2 bytes per entry, the space required is 1024 * 2 + 256 * 3 * 2 (3 second-level page tables) + 64 * (38+4+3)* 2 (38 third-level page tables for data segment, 4 for stack and 3 for code segment) = 9344 bytes.
Explanation:
16 E the answer
