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Andrew [12]
3 years ago
10

An instance of an object is (Points : 2) a block of COBOL code

Computers and Technology
1 answer:
jek_recluse [69]3 years ago
7 0

Answer: One occurrence of the object

Explanation:

 The instance of the object is the one occurrence of an object in the given class. The common statement of an object with respect to the instance means that there is single occurrence of an object.

When the process run at each time it is known as instance of program given specific values and variables. An instance of the object is also known as class instance.

It is basically defined as specific realization in an object and also the object varies in different number of the ways in the object oriented programming (OOPs).

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Help! I’ll mark you brainly! Please help me.
nika2105 [10]

Answer:

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13.  Asymmetrical balance results from unequal visual weight on each side of the composition.

14. White space

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17. The Rule of Thirds is the process of dividing an image into thirds, using two horizontal and two vertical lines.

Explanation:

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2 years ago
Suppose company A wants to develop a program that duplicates the functionality of a programm by company B. Describe how company
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Read 2 more answers
In the Unified Process (UP), related activities are grouped into UP ____. a. Services b. Disciplines c. Cycles d. Practices
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Answer:

b. Disciplines

Explanation:

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2- There are many different design parameters that are important to a cache’s overall performance. Below are listed parameters f
katen-ka-za [31]

Answer:

1. 2588672 bits

2. 4308992 bits

3. The larger the data size of the cache, the larger the area of ​​memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.

Explanation:

1. Number of bits in the first cache

Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))

total bits = 2^15 (1+14+(32*2^1)) = 2588672 bits

2. Number of bits in the Cache with 16 word blocks

Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))

total bits = 2^13(1 +13+(32*2^4)) = 4308992 bits

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