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3 years ago

A wiki can contribute to an academic paper in all of the following ways except

Computers and Technology

2 answers:

statuscvo [17]3 years ago

8 0

A. Acting as a primary source

OverLord2011 [107]3 years ago

7 0

Answer:

C.) Offering useful information

Explanation:

It's more like a 3rd-Generation source as for A and B. For D, it can, but it does that more with fiction.

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Write a Java method to delete a record from a B-tree of order n.

max2010maxim [7]

Answer:

Deleting a record on a B-tree consists of three main events:

- Searching the node where the key to be deleted exists

- Deleting the key

- Balancing the tree if required

Explanation:

q = NULL;

p = tree;

while (p) {

i = nodesearch(p, key);

q = p;

if (i < used(p) -1 && key == k(p,i)) {

found = TRUE;

position = i;

break;

}

p = son(p,i);

}

if (!found)

else if (subtree(p)) {

if (used(p) > ((n-1)/2)+1)

delkey (p, position, key);

else {

replace (p, position, fsucc(p));

p0 r1 p1 r2 p2 r3 ……. pn-1 rn-1 pn

q = &fsucc(p);

qpos = index of fsucc;

if (used(rbrother(p)) > ((n-1)/2)+1)

replace (q, qpos, sonsucc(q));

else

while (q && used(q) < (n-1)/2) {

concatenate(q, brother(q));

q = father(q);

}

else

delkey(p, position, key);

}

6 0

3 years ago

A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st

Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

$^nC_r = \frac{n!}{(n-r)!r!}$

Where $n = 15\ and\ r = 4$

$^{15}C_4 = \frac{15!}{(15-4)!4!}$

$^{15}C_4 = \frac{15!}{11!4!}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{32760}{24}$

$^{15}C_4 = 1365$

Hence, there are 1365 ways 

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

From the given parameters; 3 out of 15 is detective;

So;

$p = 3/15$

$p = 0.2$

$q = 1 - p$

$q = 1 - 0.2$

$q = 0.8$

Solving further using binomial;

$(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n$

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

$Probability = ^nC_3pq^3$

Substitute 4 for n, 0.2 for p and 0.8 for q

$Probability = ^4C_3 * 0.2 * 0.8^3$

$Probability = \frac{4!}{3!1!} * 0.2 * 0.8^3$

$Probability = 4 * 0.2 * 0.8^3$

$Probability = 0.4096$

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

Probability that at least one is defective + Probability that at none is defective = 1

Probability that none is defective is calculated as thus;

$Probability = q^n$

Substitute 4 for n and 0.8 for q

$Probability = 0.8^4$

$Probability = 0.4096$

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

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3 years ago

Who is the manager most likely to hire?

lakkis [162]

What is the question

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By default, after how much time has elapsed in a client's DHCP lease will the client attempt to renew the lease?

maw [93]

The client will attempt to renew halfway through the time of the lease.

3 0

3 years ago

Which is better computer science or software engineering?

mina [271]

you have to be extremely talented to be a software engineer so I see that as way more complex and cool.

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3 years ago

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