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3 years ago

What is a printed copy of computer output called

Computers and Technology

2 answers:

salantis [7]3 years ago

5 0

Answer:Touchscreen

Explanation:

Vsevolod [243]3 years ago

4 0

Answer:

touch screen, hard copy

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I have been trying to work on this for a while now, and this is on excel

cluponka [151]

Answer:

=IF(INDEX(INDIRECT(C10), B10)=0,"",INDEX(INDIRECT(C10), B10))

=IF((INDEX((INDIRECT(Dept,TRUE)),Reason))="","",(INDEX((INDIRECT(Dept,TRUE)),Reason)))

Explanation:

Given

Your formula:.

=INDEX(INDIRECT("Dept"),Reason)

Dept column = C (C10)

Reason field = B (B10)

The issue with your formula is that you failed to include a statement to test the falsity of the first condition; in other words, if your if statement is not true, what else should the formula do.

The question says that

"Nest the function inside an IF function so that issues currently displaying as a 0 will display as a blank cell" this means that

if the INDEX() function returns 0, a blank should be displayed in H10 blank, instead.

So, the right formula both of these two. You can use any of them

1. =IF(INDEX(INDIRECT(C10), B10)=0,"",INDEX(INDIRECT(C10), B10))

2. =IF((INDEX((INDIRECT(Dept,TRUE)),Reason))="","",(INDEX((INDIRECT(Dept,TRUE)),Reason)))

The two does the same function; the only difference is that

(1) considers the cell itself while (2) considers the contents of the cell.

The analysis of both is that

They both use a nested indirect reference to check for the content of cells displaying 0.

The first if checks for the above mentioned; if yes, cell H10 is made to display a blank else it's original content is displayed.

8 0

3 years ago

Weak Induction

slega [8]

Answer:

Following are the answer to this question:

Explanation:

In option 1:

The value of n is= 7, which is (base case)

$\to 3^7$

when n=k for the true condition:

$\to 3^k$

when n=k+1 it tests the value:

$\to 3^{(k+1)}= 3^k,3\\\to < (k!) 3 \ substituting \ equation \\\to$

since k>6 hence the value is KH>3 hence proved.

In option 2:

when:

for n=1:(base case)

$\log(1!)$

0<=0 \\ condition is true

when the above statement holds value n=1

when n=k

$\log(k!)$

when n=k+1

$\log(k+1)!=\log(k!)+\log(k+1)\\$

$\because k \log k$ $[\therefore KH>K \Rightarrow \log(KH>\loK)]$

In option 3:

when n=1:

$A_1 \cup B=A_1 \cup B\\\\$

when n=k

$\to (A_1\cap A_2 \cap.....A_k) \cup B\\=(A_1\cup B) \cap(A_2\cup B_2)....(A_k \capB).....(a)\\\to n= k+1\\ \to (A_1\cap A_2 \cap.....A_{kH}) \cup B= (A_1\cup B)\\\\\to [(A_1\cap A_2 \cap.....A_{k}) \cup B]\cap (A_{KH}\cup B)\\\\\to [(A_1\cup B) \cap (A_2 \cup B) \cap (A_3\cup B).....(A_k\cup B)\cap (A_{k+1} \cup B)\\\\ \ \ \ \ \ \ substituting \ equation \ a \\\\$