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3 years ago

When adding cells you must use a "+" symbol, you cannot use a ":" symbol. ☐ True ☐ False

Computers and Technology

1 answer:

enot [183]3 years ago

6 0

Answer:

False

Explanation:

We can use both. If the total number of cells are less we can use + and add like =Sum(A1 + A2 +A3) and if cells are in sequence, and in good number then we can use ":" like =Sum(A1:A7).

With + we can only add two cells at a time, and we need to use it again and again to add all cells, like:

If we want to add in between C1 to H1, then with plus we need to write =SUM(C1+D1+E1+F1+G1+H1).

However with : we only need to write =SUM(C1:H1), and that's it.

You might be interested in

Which of the following method calls are legal?a. mathMethod(x);b. mathMethod(y);c. mathMethod(x, y);d. mathMethod(x + y);e. math

Roman55 [17]

Answer:

Option a, b, d, f, and j is legal method calls.

Explanation:

In the given question some information is missing, that is the method definition which can be described as follows:

Method definition:

int x =2, y=3; //defining integer variable

int mathMethod (int x) //method definition

{//method body

int z=x+y; //calculate value

return z; //return value

}

In the given question option a, b, d, f, and j is legal, because these options follow the correct syntax, and other option were incorrect which can be defined as follows:

In option c, It is illegal because it accepts only one parameter.
In option e, It is accepts long number, that's why it is illegal.
In option g, It accepts float value, that's why it is illegal.
In option h, It doesn't accepts any parameter.
In option i, It isn't use in method.

7 0

3 years ago

If wire rope guardrails are used, what also must be done?

inn [45]

Some sort of security pole to hold the 2 ends together firmly <span />

6 0

3 years ago

Read 2 more answers

Write a Python program that uses function(s) for writing to and reading from a file:

swat32

import random

def random_number_file_writer(nums):

f = open("random.txt", "w")

i = 0

while i < nums:

f.write(str(random.randint(1,500))+"\n")

i += 1

f.close()

def random_number_file_reader():

f = open("random.txt", "r")

total = 0

count = 0

for x in f.readlines():

total += int(x)

count += 1

print("The total of the numbers is "+str(total))

print("The number of random numbers read from the file is "+str(count))

def main():

random_number_file_writer(int(input("How many random numbers do you want to generate? ")))

random_number_file_reader()

main()

I hope this helps!

6 0

3 years ago

25 pts! If programmers cannot write an algorithm because the problem is too complex, which methodology can they use to reach an

Setler [38]

Answer:

D, most likely thats what I was taught .

3 0

3 years ago

Recall that within the ArrayBoundedQueue the front variable and the rear variable hold the indices of the elements array where t

Citrus2011 [14]

Answer:

int n = elements.length;

if(rear < n) {

rear = (rear + 1) % n;

elements[rear] = element;

rear = rear + 1;

}

Explanation:

Options are not made available; However, I'll answer the question base on the following assumptions.

Assumptions

Array name = elements

The front and the rear variable hold the current indices elements

Element to enqueue is "element" without the quotes

To enqueue means to add an element to an already existing queue or to a new queue.

But first, the queue needs to be checked if it can accept a new element or not; in other words, if it's full or not

To do this, we check: if rear < the length of the queue

If this statement is true then it means the array can accept a new element; the new element will be stored at elements[rear] and the rear will be icremented by 1 rear by 1

Else if the statement is false, then an overflow is said to have occurred and the element will not be added.

Going by the above explanation, we have

int n = elements.length;

if(rear < n) {

rear = (rear + 1) % n;

elements[rear] = element;

rear = rear + 1;

}

Additional explanation:

The first line calculates the length of the queue

The if condition on line 2 tests if the array can still accept an element or not

Let's assume this statement is true, then we move to liine 3

Line 3 determines the new position of rear.

Let's assume that n = 6 and current rear is 4.

Line 3 will produce the following result;

rear = (rear + 1) % n;

rear = (4 + 1)% 6

rear = 5%6

rear = 5.

So, the new element will be added at the 5th index

Then line 4 will be equivalent to:

elements[rear] = element;

elements[5] = element;

Meaning that the new element will be enqueued at the 5th index.

Lastly, rear is incremented by 1 to denote the new position where new element can be added.

6 0

3 years ago