Would two bits be enough to assign a unique binary number to each vowel in the English language?

Two bits would be not enough.

Four bits would be enough.

Explanation:

With two bits, only 4 unique binary numbers are available. They are:

00

01

10

11

And since there are 5 lowercase vowel letters and 5 uppercase vowel letters in the English language, making a total of 10 vowel letters, two bits will only cater for 4 of the 10 letters and as such will not be enough.

With three bits, only 8 unique binary numbers are available. They are:

000

001

010

011

100

101

110

111

Three bits will therefore only cater for 8 of the 10 letters and as such will not be enough.

With four bits however, there are 16 binary numbers available. They are:

0000

0001

0010

0011

0100

0101

0110

0111

1000

1001

1010

1011

1100

1101

1110

1111

Four bits are more than enough to cater for all the 10 vowel letters in the English language.

PS: The number of unique binary numbers that can be found in n bits is given by;

2ⁿ

So;

If we have 3 bits, number of unique binary numbers will be

2³ = 8

If we have 6 bits, number of unique binary numbers will be

2⁶ = 64

answer: internal fragmentation is the difference between the (fixed) size of a memory hole and the data within in that memory. for example, if a 1,024 byte block of memory holds 800 bytes of data, then the 224 bytes difference is considered lost to internal fragmentation.

external fragmentation occurs when the sum of memory fragments is large enough to satisfy a request, but the holes are non-contiguous and the request cannot be satisfied. this can occur when memory is dynamically allocated and freed.

explanation:

the answer is d. splog