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salantis [7]
3 years ago
10

An ___ is any person who spends times using technology at home

Computers and Technology
1 answer:
Irina-Kira [14]3 years ago
3 0
The answer to this is home user
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Find the mistakes in the following code. Not all lines contain mistakes. Each line depends on the lines preceding it. Watch out
Kazeer [188]

Answer:

Explanation:

A null pointer is pointing to nothing, if we can detect if a pointer is null, we can use an "IF" because a null pointer always going to be false, a null pointer even can block an application, because is a result of undefined behavior, it can even cause problems because the compiler can’t tell whether we mean a null pointer or the integer 0, there is a confusion in this kind of pointers.

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A reputable, world-renowned auction house uses blockchain to verify the authenticity of paintings prior to placing them up for s
REY [17]

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The ledgers are secure, shared, and incorruptible is the correct answer.

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3 years ago
With ____ editing, Word automatically displays a Paste Options button near the pasted or moved text.
harina [27]
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3 years ago
I need help with Python. The first image is the code and the 2nd is the output.
Gekata [30.6K]
The issue arises because the string you are trying to print is not a string, rather a float value. Item1, item2 and item3 are strong values (if you type some alphabets in it and not just numbers), but itemonecost, itemtwocost, and itemthreecost are explicitly type casted to float. In line 22, 23, and 24 you’re trying to print a float, by adding it with the string. One cannot add numbers to string. Rather you can type cast the itemcost to string while printing.

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4 0
3 years ago
A digital computer has a memory unit with 16 bits per word. The instruction set consists of 72 different operations. All instruc
OLga [1]

Answer:

a. 7 bits b. 9 bits c. 1 kB d. 2¹⁶ - 1

Explanation:

a. How many bits are needed for the opcode?

Since there are 72 different operations, we require the number of bits that would contain 72 different operations. So, 2ⁿ ≥ 72

72 = 64 + 8 = 2⁶ + 8

Since n must be an integer value, the closest value of n that would contain 72 different operations is n = 7. So, 2⁷ = 128

So, we require 7 bits for the opcode.

b. How many bits are left for the address part of the instruction?

bits left = bits per word - opcode bit = 16 - 7 = 9 bits

c. What is the maximum allowable size for memory?

Since there are going to be 2⁹ bits to addresses each word and 16 bits  for each word, the maximum allowable size for memory is thus 2⁹ × 16 = 512 × 16 = 8192 bits.

We convert this to bytes

8192 bits × 1 byte/8 bits = 1024 bytes = 1 kB

d. What is the largest unsigned binary number that can be accommodated in one word of memory?

Since the number go from 0 to 2¹⁶, the largest unsigned binary number that can be accommodated in one word of memory is thus

2¹⁶ - 1

6 0
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