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never [62]
3 years ago
15

Together with some anthropologists, you’re studying a sparsely populated region of a rainforest, where 50 farmers live along a 5

0-mile-long stretch of river. Each farmer lives on a tract of land that occupies a 1-mile stretch of the river bank, so their tracts exactly divide up the 50 miles of river bank that they collectively cover. (The numbers are chosen to be simple and to make the story easy to describe.)
The farmers all know each other, and after interviewing them, you’ve discovered that each farmer is friends with all the other farmers that live at most 20 miles from him or her, and is enemies with all the farmers that live more than 20 miles from him or her.

Computers and Technology
1 answer:
Varvara68 [4.7K]3 years ago
6 0

Answer:

Explanation:

Farmers are always both directly and indirectly connected to each other

Their network is mostly strong

Networks become weak only on the edges (ends) of the river but doesn't completely dimnish

With the available network length, the center of river bank forms the strongest network of all and becomes a key player in defining the balance property of overall network

The network is very well structurally balanced and we can see that through the below image

20 miles 10 20 30 40 50

See attachment file for diagram

Considering the total length of river as 50miles and and the center of the whole length will be at 25th mile. From that point, if we consider a farmer will be be having friends for a length of 20miles both along upstream and downstream.

By this he'll be in friend with people who are around 80% of the total population. As me move from this point the integrity increases and this results in a highly balanced structural network.

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Illustrate that the system is in a safe state by demonstrating an order in which the threads may complete.If a request from thre
Kamila [148]

Answer:

a. safe sequence is T2 , T3, T0, T1, T4.

b. As request(T4) = Available, so the request can be granted immediately

c. As request(T2) < Available, so the request can be granted immediately

d. As request(T3) < Available, so the request can be granted immediately.

Explanation:

It will require matrix

[i, j] = Max [i, j] – Allocation [i, j]

A B C D

T0 3 3 3 2

T1 2 1 3 0

T2 0 1 2 0

T3 2 2 2 2

T4 3 4 5 4

Available = (2 2 2 4)

1. Need(T2) < Available so, T2 can take all resources

Available = (2 2 2 4) + (2 4 1 3) (Allocation of T2) = (4 6 3 7)

2. Need(T3)<Available so, T3 will go next

Available = (4 6 3 7) + (4 1 1 0) = (8 7 4 7)

Like wise next T0, T1, T4 will get resources.

So safe sequence is T2 , T3, T0, T1, T4.

(Note, there may be more than one safe sequence).

Solution b.

Request from T4 is (2 2 2 4) and Available is (2 2 2 4)

As request(T4) = Available, so the request can be granted immediately.

Solution c.

Request from T2 is (0 1 1 0) and Available is (2 2 2 4)

As request(T2) < Available, so the request can be granted immediately.

Solution d.

Request from T3 is (2 2 1 2) and Available is (2 2 2 4)

As request(T3) < Available, so the request can be granted immediately.

5 0
3 years ago
MCQ: A computer network is interconnection of two or more:
timama [110]

Answer:

computers

Explanation:

it is because other remaining are not able to access network

4 0
2 years ago
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snow_tiger [21]
The password would be Apple 123
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2 years ago
Think of a simple software project that requires your design skills. Specify the requirements needed on each design process
andrew-mc [135]

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Explanation:

4 0
2 years ago
We need ____ pointers to build a linked list.
adelina 88 [10]

Answer:

two

Explanation:

A linked list is a data structure which stores the multiple-element with data type

and a pointer that stores the address of the next element.

A linked list is a series of nodes connecting each other by a pointer.

a node contains data and a pointer.

For build an array, two pointers are used:

the first pointer for specifies the starting node called head node.

and the second pointer is used to connect the other node to build the linked list.

Both are used to build the array if we lose the head node we cannot apply the operation because we do not know the starting node and we cannot traverse the whole linked list.

for example:

1->2->3->4->5

here, 1 is the head node and -> denote the link which makes by the second pointer.

3 0
3 years ago
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