True, a retractable service pit cover can help other shop employees from falling into the pit while another technician is in the pit servicing a car.
Retractable service pit cover is a special cover made to cover service pit which aid inspection and repair of vehicle in an auto workshop.
Retractable service pit ensure safety of workshop technicians around service pits.Retractable service pits aid proper inspection of the beneath of the vehicleRetractable service pits enable the technician to have a close view of the vehicle in repair.The pit also serves as guide to short-eyed people or people to are not aware of the structure.In essence, the pit prevent against injury or death of workers and occupier.In conclusion, the retractable service pit cover help other shop employees from falling into the pit even if they are aware of the structure.
Learn more about retractable service pit in picture attached
"Test Phase " is the correct choice.
Explanation:
DevSecOps seems to be a community as well as experience of corporate data science which encompasses software design, regulation, including operational activities. This same main feature of DevSecOps has always been to strengthen customer achievement as well as expedition importance by computerizing, supervising as well as implementing data protection at all stages of the development including its development tools.The testing method throughout the test phase would then help make sure that the controller is designed mostly under the responsibilities forecasted. The test focuses on either the reaction times, dependability, use of resources but instead interoperability of applications.NTQ
Explanation:
The given sequence is
BHE : FLI : JPM
If is clear that, alphabets on first places are B, F, J. Difference between their place vales is 4.
B+4=F,F+4=J ; so alphabet on first place of next term of sequence is J+4=N.
Similarly, alphabets on second places are H, L, P. Difference between their place vales is 4.
H+4=L,L+4=P ; so alphabet on second place of next term of sequence is P+4=T.
Alphabets on third places are E, I, M. Difference between their place vales is 4.
E+4=I,I+4=M ; so alphabet on third place of next term of sequence is M+4=Q.
Therefore, the next term is NTQ.
The following code or the program will be used
Explanation:
def readFile(filename):
dict = {}
with open(filename, 'r') as infile:
lines = infile.readlines()
for index in range(0, len(lines) - 1, 2):
if lines[index].strip()=='':continue
count = int(lines[index].strip())
name = lines[index + 1].strip()
if count in dict.keys():
name_list = dict.get(count)
name_list.append(name)
name_list.sort()
else:
dict[count] = [name]
print(count,name)
return dict
def output_keys(dict, filename):
with open(filename,'w+') as outfile:
for key in sorted(dict.keys()):
outfile.write('{}: {}\n'.format(key,';'.join(dict.get(key
print('{}: {}\n'.format(key,';'.join(dict.get(key
def output_titles(dict, filename):
titles = []
for title in dict.values():
titles.extend(title)
with open(filename,'w+') as outfile:
for title in sorted(titles):
outfile.write('{}\n'.format(title))
print(title)
def main():
filename = input('Enter input file name: ')
dict = readFile(filename)
if dict is None:
print('Error: Invalid file name provided: {}'.format(filename))
return
print(dict)
output_filename_1 ='output_keys.txt'
output_filename_2 ='output_titles.txt'
output_keys(dict,output_filename_1)
output_titles(dict,output_filename_2)
main()
x=float(input("Enter a number: "))
sub=(x-int(x))
print(sub)
Explanation:
Got it Right
understanding others first before starting to explain your own point
Explanation:
Stephen covey believes the key to effective communication is understanding others first before starting to explain your own point. Communication is an important skill in life, in which one must have a good character in order to communicate effectively.
len(word2) >= len(word1) and len(word2) >= len(word3):
Question with blank is below
def longest_word(word1,word2,word3):
if len(word1) >= len(word2) and len(word1) >= len(word3):
word = word1
elif _________________________________________
word = word2
else:
word = word3
return word
print(longest_word("chair","couch","table"))
print(longest_word("bed","bath","beyond"))
print(longest_word("laptop","notebook","desktop"))
print(longest_word("hi","cat","Cow"))
Explanation
In line 1 of the code word1, word2, and word3 are the parameters used to for the defining the longest_word function. They will be replaced by 3 words to be compared. The code that is filled in the blank is len(word2) >= len(word1) and len(word2) >= len(word3): It is a conditional statement that is true only if the number of characters in the string of word2 is greater than or equal to word1 and word2 is greater than that of word3 .
Here is the fractional_part() function:
def fractional_part(numerator, denominator):
if denominator != 0:
return (numerator % denominator)/denominator
else:
return 0
Explanation:
I will explain the code line by line.
The first statement it the definition of function fractional_part() which takes two parameters i.e. numerator and denominator to return the fractional part of the division.
Next is an if statement which checks if the value of denominator is 0. If this is true then the function returns 0. If this condition evaluates to false which means that the value of denominator is not 0 then return (numerator % denominator)/denominator is executed. Now lets see how this statement works with the help of an example.
Lets say the value of numerator is 5 and denominator is 4. (numerator % denominator)/denominator will first compute the modulus of these two values. 5 % 4 is 1 because when 5 is divided by 4 , then the remainder is 1. Now this result is divided by denominator to get the fractional part. When 1 is divided by 4 the answer is 0.25. So this is how we get the fractional part which is 0.25.
The program with output is attached.
It will provide an instant answer!