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2 years ago

Select the described workplace culture from the drop-down menu.

Computers and Technology

2 answers:

Vilka [71]2 years ago

7 0

Answer:

Adaptive
Consistency-oriented
Achievement-based
Involvement-oriented

Explanation:

Adaptive Consistency-oriented workplace cultures emphasizes reacting quickly to change.

Consistency-oriented workplace cultures emphasizes everyone being in agreement.

Achievement-based Consistency-oriented workplace cultures emphasizes accomplishing goals.

Involvement-oriented Achievement-based workplace cultures emphasize development of new skills.

Ostrovityanka [42]2 years ago

5 0

Answer:

Adaptive Consistency-oriented workplace cultures emphasizes reacting quickly to change.

Consistency-oriented workplace cultures emphasizes everyone being in agreement.

Achievement-based Consistency-oriented workplace cultures emphasizes accomplishing goals.

Involvement-oriented Achievement-based workplace cultures emphasize development of new skills.

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What conclusion can be made about the state of the program when the while loop terminates? Assume answer is a declared and initi

stellarik [79]

Answer:

1. The loop will run unless the value of answer equals to Capital N. The method equals is “case-sensitive” that is lower case letter and upper case letters are considered different. That “a” and “A” means the same to us, but for the method equals it is different and equals method will return false if the case is not matching.

2. a (int)(Math.random() * (upper − lower) ) + lower 

Since we need to consider the lower limit value together to get the desired results we need to add the value of lower limit to the multiplied answer.

3. b while(userGuess != secretNumber || userGuess >= lowerLimit && userGuess <= upperLimit)

Here the program is required to check the while loop and exit when user guess the number or we can say the loop should continue until the user guess the number, so that is why we have taken userGuess!=secretNumber. Next the loop should be exited if it is not within the range or we can say that the loop should run only if the guessed number is within the upper and lower limit. That is why we have opted for the condition userGues>=lowerlimit && userGuess<=upperlimit. Why we have taken && as the operator is that, it’s a range of values so && operator best suit for this kind of logical condition.

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2 years ago

Show how to define a view tot_credits (year, num_credits), giving the total number of credits taken by students in each year

andreev551 [17]

Hi! I'm a Digital Marketer Intern at hotels.ng and I have a moderate knowledge on programming.

First, your question is not very explanatory. The term "view" is often used in back-end web development. A view is simply a Python function that takes a Web request and returns a Web response.

But I'm not sure this is what you want, so I'll just go ahead and write a python function involving class to return the total number of credits taken by a student.

I'll answer this question using Python.

class student(object):

credits = None

year = None

def num_credits(self):

#get credit value

self.credits = input("Enter the total number of credits: " )

pass

def getYear(self):

self.year = input("Enter current year: ")

pass

def tot_credits(self):

TotalCredits = tempTotalCredits + self.num_credits

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2 years ago

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Dafna1 [17]

Answer:

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Explanation:

Microsoft Excel provides the power of tables and list for categorized data and data validation.

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2 years ago

Use the loop invariant (I) to show that the code below correctly computes the product of all elements in an array A of n integer

NeTakaya

Answer:

Given Loop Variant P = a[0], a[1] ... a[i]

It is product of n terms in array

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The Basic Step: i = 0, loop invariant p=a[0], it is true because of 'p' initialized as a[0].

Induction Step: Assume that for i = n - 3, loop invariant p is product of a[0], a[1], a[2] .... a[n - 3].

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And multiply a[n-1] with p, i.e P = a[0].a[1].a[2].... a[n-2].

a[n-1]. i.e. P=P.a[n-1]

By the assumption for i=n-3 loop invariant is true, therefore for i=n-2 also it is true.

By induction method proved that for all n > = 1 Code will return product of n array elements.

While loop check the condition i < n - 1. therefore the conditional statement is n - i > 1

If i = n , n - i = 0 , it will violate condition of while loop, so, the while loop will terminate at i = n at this time loop invariant P = a[0].a[1].a[2]....a[n-2].a[n-1]

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