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Nataliya [291]
3 years ago
11

USDA-APHIS Animal Care Resource Policy #12 "Considerations of Alternatives to Painful/Distressful Procedures" states that when a

database search is the primary means of considerations to alternatives of painful and distressful procedures, that a narrative should include all of the following except:
A. The name of the databases searched
B. The date the search was performed
C. The specific field of study
D. The search strategy (including scientifically relevant terminology) used
Click card to see definition ????
C. The specific field of study
Click again to see term ????
Which of the following should be considered a significant change to an animal use proposal?
A. change in departmental affiliation of the principal investigator
B. change in start date of study
C. change in anesthetic agent
D. change in fluid administration
Click card to see definition ????
C. change in anesthetic agent
Click again to see term ????
1/566



Created by
edgar_rowton
Terms in this set (566)
USDA-APHIS Animal Care Resource Policy #12 "Considerations of Alternatives to Painful/Distressful Procedures" states that when a database search is the primary means of considerations to alternatives of painful and distressful procedures, that a narrative should include all of the following except:
A. The name of the databases searched
B. The date the search was performed
C. The specific field of study
D. The search strategy (including scientifically relevant terminology) used
Computers and Technology
1 answer:
katrin2010 [14]3 years ago
6 0

Answer:

C. The specific field of study

C. change in anesthetic agent

Explanation:

In the consideration for the procedure, some information are required and examples are the strategy of the search/the database name. However, the area of study is not one of the information.

The animal use proposal can general be altered based on the type of change required. However, an anesthetic agent alteration is not a significant change.

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Implement a Python function with the signature Transfer(S, T) that transfers all elements from the ArrayStack instance S onto th
Mashutka [201]

Answer:

Following are the program in the Python Programming Language.

#define function

def Transfer(S, T):

 #set for loop  

 for i in range(len(S)):

   #append in the list

   T.append(S.pop())

 #return the value of the list

 return T

#set list type variable

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#print the values of the list

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#set the list empty type variable

T=[]

#call the function

T = Transfer(S, T)

#print the value of T

print(T)

<u>Output:</u>

['a', 'b', 'c', 'd']

['d', 'c', 'b', 'a']

Explanation:

Here, we define the function "Transfer()" in which we pass two list type arguments "S" and "T".

  • Set the for loop to append the values in the list.
  • Then, we append the value of the variable "S" in the variable "T".
  • Return the value of the list variable "T" and close the function.
  • Then, set the list data type variable "S" and initialize the elements in it and print that variable.
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Write a program that keeps track of a simple inventory for a store. While there are still items left in the inventory, ask the u
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Explanation:

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It takes 2 seconds to read or write one block from/to disk and it also takes 1 second of CPU time to merge one block of records.
Alexxx [7]

Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way  merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

<em>For optimal 4-way merging, we need one dummy run with size 0.</em>

  1. Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300
  2. Merge the run as made in step 1 with the remaining 3 runs bearing length 1000, 1200, 1500. The merged run length is 6000 and is calculated as follows

       L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000

<em>The resulting run has length 6000 records</em>.

Part b

<u><em>For step 1</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 2300 for step 01
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So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec

So the input/output time is 46 seconds for step 01.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec

So the CPU  time is 23 seconds for step 01.

Total time in step 01

T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\

Total time in step 01 is 69 seconds.

<u><em>For step 2</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec

So the input/output time is 120 seconds for step 02.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec

So the CPU  time is 60 seconds for step 02.

Total time in step 02

T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\

Total time in step 02 is 180 seconds

Merging Time (Total)

<em>Now  the total time for merging is given as </em>

T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\

Total time in merging is 249 seconds seconds

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