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stiv31 [10]
3 years ago
8

Define the following BASIC terms: 1. Keywords 2. Constants 3.Variables

Computers and Technology
1 answer:
Mamont248 [21]3 years ago
8 0

Answer:

Explanation:

  1. Keywords: Keywords are those words which have special meanings in QBASIC. Keywords are formed by using characters of QBASIC characters set. Keywords are statements, commands, functions (built in functions) and names of operators. The keywords are also called Reserved Words. Some reserved words are CLS, REM, INPUT, LET, PRINT, FOR, DO, SELECT, ASC, SQR, LEN, and INT.
  2. Constants: They are divided into:
  • Numeric constants: include all numbers (real, integer, not real,...)
  • Character constants: include all character sete (letters, digits, symbols) between two double quotation marks e.g. "BASIC", "the width is 83" etc.

   3. Variables: A variable is a name which can hold or contain a value.

Hope this helps!!

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The average number of entities completed per unit time -- the output rate -- from a process is called ____.
Sergio039 [100]

Answer:

Throughput.

Explanation:

The average number of entities completed per unit time -- the output rate -- from a process is called throughput.

Basically, throughput is a technical term used to describe the amount of data that can be transmitted or transferred from one place to another in a specific period of time. It is used to measure or determine the performance of RAM and hard drives.

8 0
2 years ago
Array A is not a heap. Clearly explain why does above tree not a heap? b) Using build heap procedure discussed in the class, con
Aleksandr [31]

Answer:

Sorted Array A { } = { 1, 4, 23, 32, 34, 34, 67, 78, 89, 100 }

Explanation:

Binary tree is drawn given that the binary tree do not follow both minimum heap and maximum heap property, therefore, it is not a heap.

See attached picture.

6 0
3 years ago
There are 55 registers, and total 60 instructions available in a general-purpose computer. The computer allows only 2-address in
Elenna [48]

Answer:

total bits = 6 + 6 + 19 = 31 bits

Explanation:

given data

total registers = 55

memory  size = 64 KB

total instructions = 60

solution

here we have given 55 register so we get greater or equal power of 2 that is here 64

2^6 = 64

so here for register operand 6 bit is required

and

when instruction 60 we get here greater or equal power of 2 that is here 64  

2^6 = 64

so here also  for represent instruction 6 bit is required

and

for size 64 kb

2^6 * 2^{10} * 2^3 = 2^{19}

so 19 bits is required for memory location

and

as instruction in 2 parts are opcode and operand

and here given as 2 address instruction

they are memory operand and the register operand

so here

total bits will be = opcode + register operand + memory operand

total bits = 6 + 6 + 19 = 31 bits

total bits =  31 bits

5 0
3 years ago
The term “computer literacy” dates back to what decade? <br> 1960s<br> 1970s<br> 1980s<br> 1990s
NNADVOKAT [17]
1970-80s which is when some of the first computers were created like apple computers which looked like a giant cube and a rectangle on the side which is very different than what we have today.

Hope this helps!
3 0
3 years ago
What is the utility of a lever?
Kamila [148]
Muy cosas.
 fazer una montanha russa andar,ativar una armadilha, ativar una puerta etc.
3 0
3 years ago
Read 2 more answers
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