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1 year ago

Determine the value of X if (211) base x = (152) base 8, how to do this?

1 answer:

Alex1 year ago

7 0

we translate the following statement given in terms of logarithms. 211 base x can be expressed into log 211 / log x while 152 base 8 can be expressed int o log 152 over log 8. In this case,
log 211 / log x = log 152 / log 8log x = 0.962x = 10^0.962 = 9.1632

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What are the advantages and disadvantages of using a very large memory cell size, say, W = 64 instead of the standard size W = 8

stepladder [879]

Answer and Explanation:

The benefits of an large size memory cell is that you can store a lot bigger numbers (and all information is, eventually, numbers) in one cell, which by and large improves the speed and unwavering quality of information get to.

The inconvenience is that you go through a bigger bit of whatever information stockpiling medium you are utilizing on each expression of information, and accordingly can store less information generally speaking.

The biggest positive integer that could be put away in a framework utilizing sign/size documentation, with 64-piece cells is 263. On the off chance that two such cells were utilized to store numbers, the biggest whole number that could be put away by the framework is 2127.

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1 year ago

Two devices are connected to the Internet and communicating with one another. A squirrel chews through one of the wires in the n

bearhunter [10]

Answer:

Fault-Tolerance

Explanation:

Fault tolerance refers to the ability of a system (computer, network, cloud cluster, etc.) to continue operating without interruption when one or more of its components fail.

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2 years ago

Develop an sec (single error correction) code for a 16-bit data word. generate the code for the data word 0101000000111001. show

Kipish [7]

Answer:

code = 010100000001101000101

Explanation:

Steps:

The inequality yields $2^{k} - 1 > = M+K$ , where M = 16. Therefore,

The second step will be to arrange the data bits and check the bits. This will be as follows:

Bit position number Check bits Data Bits

21 10101

20 10100

The bits are checked up to bit position 1

Thus, the code is 010100000001101000101

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2 years ago

Read 2 more answers

2 Name the package that contains scanner class?

Marat540 [252]

the answer is Java.util.scanner

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1 year ago

Consider a disk that rotates at 3600 rpm. The seek time to move the head between adjacent tracks is 2 ms. There are 32 sectors p

jasenka [17]

Answer:

19.71 ms

Explanation:

The disk rotates at 3600 rpm, hence the time for one revolution = 60 / 3600 rpm = 16.67 ms.

Hence time to read or write on a sector = time for one revolution / number of sectors per track = 16.67 ms / 32 sectors = 0.52 ms

Head movement time from track 8 to track 9 = seek time = 2 ms

rotation time to head up sector 1 on track 8 to sector 1 on track 9 = 16.67 * 31/32 = 16.15 ms

The total time = sector read time +head movement time + rotational delay + sector write time = 0.52 ms + 2 ms + 16.15 ms + 0.52 ms = 19.19 ms

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1 year ago