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What else can I help you with?
How many cents is 6 bits?
2 bits equal 25 cents. So 6 bits would be 75 cents.
How many bits of data are represented by a single baud in CAP-64?
6 bits
How many combinations can be formed from 6 bits?
26 = 64
How many bits are necessary to assign addresses to 62 hosts?
6
How many bits would be needed to represent a 45 character set?
45 in binary is 101101, so you need at least 6 bits to represent 45 characters.
How many bits are needed to represent the faces on a pair of dice?
A standard die has 6 faces, so to represent the faces on a single die, you need 3 bits (since 2^3 = 8, which can cover the 6 faces). For a pair of dice, you have 2 dice, which means you need 3 bits for each die, resulting in a total of 6 bits (3 bits for the first die and 3 bits for the second die). Therefore, 6 bits are required to represent the faces on a pair of dice.
How many host bits are necessary to assign addresses to 62 hosts?
6
How many different values can be represented by 6 bits 7 bits 8 bits and 10 bits?
To find out how many different values can represented by a certain number of bits, we can use the following formula 2n-1 and that is because the first number is always a zero.Based on that 6 bits = 26- 1= 64-1=637 bits= 27-1= 1278 bits= 28-1=25510 bits= 210-1=1023# of bits1=12=33=74=155=316=637=1278=2559=51110=1023
How many money in acconut no 1015949-6?
how many money in account no 101-5949-6
How many bits are in 172.168.2.0 26 subnet?
The subnet mask for a /26 subnet indicates that the first 26 bits are used for the network portion, leaving 6 bits for the host portion. In this case, the subnet 172.168.2.0/26 can accommodate 2^6 = 64 addresses, but only 62 are usable for hosts (subtracting the network and broadcast addresses). Thus, there are 26 bits designated for the network and 6 bits for hosts within this subnet.
What is a 6Mbps?
6 mega bits per second (equavalent of 6 million bits per second). If this is the speed of your internet, it is raesonable, but far from the fastest. Many countries now will offer connections at 1000 Mbps.
Consider a logical address space with 4 pages of 1024 bytes per pageeach mapped onto a physical memory of 64 frames. how many bits are there in logical address?
4 pages -> 2^2 bits 1024 bytes -> 2^10 bits 64 frames -> 2^6 bits Therefore: Logical memory = 2+10=12 bits Physical memory = 10 +6 =16 bits
