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3 years ago

What is the other name for repetitive strain injury

Computers and Technology

2 answers:

SSSSS [86.1K]3 years ago

5 0

work-related upper limb disorder, or non-specific upper limb pain.

Ostrovityanka [42]3 years ago

4 0

Carpel Tunnel syndrome, is what it is called in the USA

Hope it helped!

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Which of the following sentences uses correct punctuation? A. I am a good communicator, and I am a strong team member. B. I comm

adelina 88 [10]

A has a comma then and so that isn't correct
B should be and I am
D should have an and in it
C is correct

7 0

3 years ago

Read 2 more answers

Hello everyone can help me give all the answer

krek1111 [17]

Answer:

1. were

2. were

3 was

4 was

5 were

1 wasnt

2 werent

3werent

4 wasnt

5 werent

were you at work?

was it in the garden?

were they worried?

was lucy present?

were his friends late?

1 they were

2 it was

3 he wasn't

4 I was

5 they weren't

Explanation:

7 0

3 years ago

Consider a system consisting of m resources of the same type, being shared by n processes. Resources can be requested and releas

Mama L [17]

Answer:

Explanation:

The system will be deadlock free if the below two conditions holds :

Proof below:

Suppose N = Summation of all Need(i), A = Addition of all Allocation(i), M = Addition of all Max(i). Use contradiction to prove.

Suppose this system isn't deadlock free. If a deadlock state exists, then A = m due to the fact that there's only one kind of resource and resources can be requested and released only one at a time.

Condition B, N + A equals M < m + n. Equals N + m < m + n. And we get N < n. It means that at least one process i that Need(i) = 0.

Condition A, Pi can let out at least 1 resource. So there will be n-1 processes sharing m resources now, Condition a and b still hold. In respect to the argument, No process will wait forever or permanently, so there's no deadlock.

5 0

3 years ago

(a) How many locations of memory can you address with 12-bit memory address? (b) How many bits are required to address a 2-Mega-

I am Lyosha [343]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Let,

The address of 1-bit memory to add in 2 location:

$\to \frac{0}{1} =2^1 \ (\frac{m}{m} \ location)$

The address of 2-bit memory to add in 4 location:

$\to \frac{\frac{00}{01}}{\frac{10}{11}} =2^2 \ (\frac{m}{m} \ location)$

similarly,

Complete 'n'-bit memory address' location number is = $2^n.$ Here, 12-bit memory address, i.e. n = 12, hence the numeral. of the addressable locations of the memory:

$= 2^n \\\\ = 2^{12} \\\\ = 4096$

In point b:

$\to Let \ Mega= 10^6$

$=10^3\times 10^3\\\\= 2^{10} \times 2^{10}$

So,

$\to 2 \ Mega =2 \times 2^{20}$

$= 2^1 \times 2^{20}\\\\= 2^{21}$

The memory position for ' $2^n$ ' could be 'n' m bits'

It can use $2^{21}$ bits to address the memory location of 21.

That is to say, the 2-mega-location memory needs '21' bits.

Memory Length = 21 bit Address

In point c:

$i^{th}$ element array addresses are given by:

$\to address [i] = B+w \times (i-LB)$

$_{where}, \\\\B = \text {Base address}\\w= \text{size of the element}\\L B = \text{lower array bound}$

$\to B=\$ 52\\\to w= 4 byte\\ \to L B= 0\\\to address = 10$

$\to address [10] = \$ 52 + 4 \times (10-0)\\$

$= \$ 52 + 40 \ bytes\\$

1 term is 4 bytes in 'MIPS,' that is:

$= \$ 52 + 10 \ words\\\\ = \$ 512$

In point d:

$\to base \ address = \$ t 5$

When MIPS is 1 word which equals to 32 bit :

In Unicode, its value is = 2 byte

In ASCII code its value is = 1 byte

both sizes are < 4 byte

Calculating address:

$\to address [5] = \$ t5 + 4 \times (5-0)\\$

$= \$ t5 + 4 \times 5\\ \\ = \$ t5 + 20 \\\\= \$ t5 + 20 \ bytes \\\\= \$ t5 + 5 \ words \\\\= \$ t 10 \ words \\\\$

3 0

3 years ago

"what type of database allows new tables to be created out of information that is requested in a search of the database

Talja [164]

If you don't put propper measures into place while creating this database your site will vulnrable to SQL Injections. In this age most sites aren't because people have learned from past mistakes. My advice do some pentesting when you finish the site and if is vulnrable to SQL Injections quit web devlopment!

3 0

3 years ago

Read 2 more answers