Answer:
for(i = 0 ; i < NUM_VALS; ++i)
{
cout << courseGrades[i] << " ";
}
cout << endl;
for(i = NUM_VALS-1 ; i >=0 ; --i)
{
cout << courseGrades[i] << " ";
}
cout << endl;
Explanation:
The first loop initializes i with 0, because we have to print the elements in order in which the appear in the array. We print each element, adding a space (" ") character at its end. After the loop ends, we add a new line using endl.
The second loop will print the values in a reverse order, so we initialize it from NUM_VALS-1, (since NUM_VALS = 4, and array indices are 0,1,2,3). We execute the loop till i >= 0, and we print the space character and new line in a similar way we executed in loop1.
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Explanation:
Answer:
a) AL will contains 0011 1100
Explanation:
In assembly language, shifting bits in registers is a common and important practice. One of the shifting operations is the SHR AL, x where the x specifies that the bits be shifted to the right by x places.
SHR AL, 2 therefore means that the bits contained in the AL should be shifted to the right by two (2) places.
For example, if the AL contains binary 1000 1111, the SHR AL, 2 operation will cause the following to happen
Original bit => | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
Shift once to the right => | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | (0) |
Shift once to the right => | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | (0) | (0) |
Notice;
(i) that there are two shifts - one at a time.
(ii) that the bits in bold face are the bits in the AL after the shift. Those that in regular face are those in the carry flag.
(iii) that the new bits added to the AL after a shift are the ones in bracket. They are always set to 0.
Answer:
The answer is "
"
Explanation:
Calculating the mean for brand A:
Calculating the Variance for brand A:
Calculating the Standard deviation:
Calculating the Mean for brand B:
Calculating the Variance for brand B:
Calculating the Standard deviation:
This is the correct Answer <span>Attribute</span>
