Nt1310 Unit 7 Lab Report

23. Consider a 4-TB disk that uses 4-KB blocks and the free-list method. How many block

The pattern that must be repeated is arranging the dots three by three (3x3) for each row horizontally and vertically, and not more than three dots diagonally.

4. The aircraft had to make two figure eights and fly over a 10 foot obstacle at the start and finish.

2.OA.C.4. Use addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns; write an equation to express the total as a sum of equal

N – (k + 1) = 30 – (4 + 1) = 25 degrees of freedom

This thin line if stretched out would measure approximately 8 kilometers. When the CD has had data stored upon it there are a series of bumps and pits within these strands as seen in figure 2. Each bump and pit is the form of data stored on the disc. The computer reads these alterations on the disc as either one or zero. Or is in the form of Binary Code

10. The number of orbitals having a given value of l is equal to A) 2n + 1 B) 2l + 1 C) n + ml D) 2ml + 1 E) l + ml

The metal layer of the CD is the layer that contains the data. CD data is represented as tiny indentations known as "pits”, these pits are encoded in a spiral track that if unrolled from it spiral would be around 5.5 km long, the spiral is moulded into the top of the polycarbonate layer, these are microscopic and usually expressed in the measurement of a micron µm which is 1 millionth of a meter. The areas between pits are known as "lands". The pits are even smaller than the lands and are approximately 100 nm deep by 500 µm wide, and vary from 850 µm to 3.5 µm in length. The measurement of the pit represents the binary information of the data stored on the disc.

D − R + = 5 6 7 8 4 3 7 8 = 4 3 8 7

Answer 4: Here, we need to move 5 large cans of paint from basement to the second floor. Let us assume that the Force for lifting one can is 1N and the height of one floor is 1m. Thus, the below two scenarios are taken as:

Answer b) Loop Invariant: At the start of each iteration (j) of the for loop (lines 2-4), A[j] contains the smallest element in the range (j … A.length). Also, the elements in the range (j… A.length) are the rearrangement of the original elements in the range (j… A.length).

Like the patterns observed in examples 6 and 7, Sn tends to ax, which is 8 in this case. An interesting pattern can be observed from the above three examples of 6-8. Sn tends to ax.

The second difference of the sequence is divided by 2, since this is the value of a, thereafter the sequence is extended to T0, as the number highlighted in red is c. To figure out the value of b, the values have been substituted into the quadratic formula as shown below, where after the equation has been solved using simple algebra.

for (startH = 0; startH < N / 2; startH = startH * 3 + 1);

9. The procedure is repeated by moving the horizontal weight this time to the left.