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kifflom [539]
2 years ago
6

The ub-04 form would most likely be used in what setting?

Computers and Technology
1 answer:
wlad13 [49]2 years ago
4 0

The UB-04 form would most likely be used in the following setting:

  • Hospitals
  • Hospices
  • General health centers such as rural clinics.
  • Rehab facilities

The uniform bill (UB-04) form is also referred to as CMS-1450 and it is a standard claim form that is used for billing and submitting Medicare Part A outpatient or inpatient health claims to Medicare fiscal intermediaries such as insurance companies, medical providers (hospitals), etc.

Also, the uniform bill (UB-04) form was introduced by the National Uniform Billing Committee (NUBC) in 2004, so as to replace the UB-92 form.

Generally, UB-92 form is a paper claim form that is typically used for laboratory, radiology, surgery, etc. Thus, the UB-04 form would most likely be used in the following setting:

  • Hospitals
  • Hospices
  • General health centers such as rural clinics.
  • Rehab facilities

Read more on UB-04 form here: brainly.com/question/4437165

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Minchanka [31]

Answer:

<u>The total time elapsed from the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B is </u><u>25.11 ms</u>

Explanation:

Host A first converts the analog signal to a digital 64kbps stream and then groups it into 56-byte packets. The time taken for this can be calculated as:

time taken 1= \frac{Packet Size in Bits}{Bit Rate}

                 = (56 x 8) bits / 64 x 10³ bits/s

                 = 7 x 10⁻³s

time taken 1= 7 ms

The transmission rate of the packet from Host A to Host B is 4 Mbps. The time taken to transfer the packets can be calculated as:

time taken 2= (56 x 8) bits / 4 x 10⁶ bits/s

                    = 1.12 x 10⁻⁴ s

time taken 2= 112 μs

The propagation delay is 18 ms.

To calculate the total time elapsed, we need to add up all the time taken at each individual stage.

<u />Time_{total}<u> = Time taken 1 + Time taken 2 + Propagation Delay</u>

                 = 7 ms + 112 μs + 18 ms

                 = 0.025112 s

Time_{total} = 25.11 ms

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In order to print multiple worksheets, what needs to happen before printing?
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Answer:

Explanation:

Click the File tab.

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Given that it takes 0.08 ms to travel from one track to the next of a hard drive; that the arm is originally positioned at Track
daser333 [38]

Answer:

Time taken to travel from one track to the next = 0.08ms

Initial track= 15     0

4      (15-4)*(0.08)= 0.88

40     (40-4)*(0.08)= 2.88

35      (40-35)*(0.08)= 0.4

11       (35-11)*(0.08)= 1.92

14       (14-11)*(0.08)= 0.24

7         (14-7)*(0.08)= 0.56

----------------------------------------------

Total seek time=0.88+2.88+0.4+1.92+0.24+0.56=6.88ms

Explanation:

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