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dalvyx [7]
3 years ago
10

01000100 01101111 00100000 01111001 01101111 01110101 00100000 01101011 01101110 01101111 01110111 00100000 01111001 01101111 01

110101 01110010 00100000 01100010 01101001 01101110 01100001 01110010 01111001 00100000 01100011 01101111 01100100 01100101 00111111 00001010 01000001 01101110 01110011 01110111 01100101 01110010 00100000 01110100 01101000 01101001 01110011 00100000 01101001 01101110 00100000 01100010 01101001 01101110 01100001 01110010 01111001 00100000 01100011 01101111 01100100 01100101
(another 30 point easy question (Yes I do extra point easy questions a lot...that's why I'm still a helping hand despite answering so many things and yes this is my last easy question of the day NO TRANSLATING)
Computers and Technology
2 answers:
dalvyx [7]3 years ago
6 0

01011001 01100101 01100001 00100000 01001001 00100000 01101011 01101110 01101111 01110111 00100000 01101101 01111001 00100000 01100010 01101001 01101110 01100001 01110010 01111001 00101100 00100000 01100100 01101111 00100000 01111001 01101111 01110101 00111111 00100000 01001100 01001111 01001100 00100000 01101010 01101011



^^^^ What do you think

Anna71 [15]3 years ago
5 0
This was quite tricky but, I found a program online that answers your question. The answer is (Do you know your binary code? Answer this in binary code)
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Convert one billion byte into one storage unit​
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Answer:

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What will the following code print out: int numbers [] = {99, 87, . 66, 55, 101}; for (int i = 1; i < 4; i++) cout << n
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the array show a decimal value .66, i assume this is enter by mistake because option has no decimal value.

The index of the array is start from zero, it means the first element store in the array at position zero.

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3 years ago
Providing incentives for customers to learn more about your service is known as?
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3 years ago
What is the exact number of bytes in system of memory that contains (a) 64K bytes (In Binary)(b) 32M bytes
VashaNatasha [74]

Answer:

64 K bytes = 65536 bytes

32 M bytes = 33554432 bytes

Explanation:

The question expect the number of bytes in binary instead of decimal. So this is important to understand that:

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Therefore,

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Using the similar calculation logic, we know

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Therefore,

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