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Anton [14]
3 years ago
10

When selecting font size in a document, ____ points equals one-inch-tall text?

Computers and Technology
1 answer:
IgorC [24]3 years ago
3 0
Hello <span>Jhigg4685 
</span><span>

Answer: When selecting font size in a document, 72 points equals one-inch-tall text?


I hope this helps
-Chris</span>
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How much memory did the first smartphone have?
hjlf

Memory 1 MB (2 × HM658512LTT PSRAM)

Storage 1 MB of NOR Flash expanded to 2 MB by Stacker compression + 32KB BIOS NOR Flash

6 0
3 years ago
In the source-code file containing a class’s member function definitions, each member function definition must be tied to the cl
jenyasd209 [6]

Answer:

Binary Scope Resolution Operator

Explanation:

The :: (scope resolution) operator is used to get hidden names due to variable scopes so that you can still use them. The scope resolution operator can be used as both unary and binary. Fortunately, in this context we're not dealing with hidden contents; thus the Binary applies better than the Unary.

Cheers

5 0
3 years ago
g Given a character and a list of strings, find strings that do not contain the given character. See the example below. $ (Find
Alex_Xolod [135]

Answer:

The solution code is written in Python 3

  1. def findStr(stringList, c):
  2.    output = []
  3.    for currentStr in stringList:
  4.        if c not in currentStr.lower():
  5.            output.append(currentStr)
  6.    return output  
  7. strList = ["Apple", "Banana", "Grape", "Orange", "Watermelon"]
  8. print(findStr(strList, "g"))

Explanation:

Firstly, we can create a function and name it as findStr which takes two input parameters stringList and a character, c (Line 1).

Create a list that will hold the list of strings that do not contain the input character (Line 2).

Create a for-loop to traverse through each string in the list (Line 3).

In the loop, check if the input character is not found in the current string (Line 4), if so, add the current string to the output list (Line 5). Please note we also need to convert the current string to lowercase as this program should ignore the casing.

After completion the loop, return the output list (Line 7).

We can test the function using a sample string list and input character "g" (Line 9 - 10). We shall get the output as follows:

['Apple', 'Banana', 'Watermelon']

5 0
3 years ago
A user has requested to have an extra 4GB of memory installed on a laptop. During the Installation process, which of the followi
9966 [12]

Answer: C) Access panel

Explanation:

 During the installation process, access panel is used remove by the technician for installation extra 4 GB memory in the laptop. As, window access panel is useful to change the window settings.

We need to access the panel in the laptop to add extra 4 GB memory in the laptop as, the access panel contain the memory banks in the system. Then, open that particular panel for upgrade the memory in the laptop and also remove the unnecessary memory.

8 0
3 years ago
(a) How many locations of memory can you address with 12-bit memory address? (b) How many bits are required to address a 2-Mega-
I am Lyosha [343]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Let,

The address of 1-bit  memory  to add in 2 location:

\to \frac{0}{1}  =2^1  \ (\frac{m}{m}  \ location)

The address of 2-bit  memory to add in 4 location:

\to \frac{\frac{00}{01}}{\frac{10}{11}}  =2^2  \ (\frac{m}{m}  \ location)

similarly,

Complete 'n'-bit memory address' location number is = 2^n.Here, 12-bit memory address, i.e. n = 12, hence the numeral. of the addressable locations of the memory:

= 2^n \\\\ = 2^{12} \\\\ = 4096

In point b:

\to Let \  Mega= 10^6

              =10^3\times 10^3\\\\= 2^{10} \times 2^{10}

So,

\to 2 \ Mega =2 \times 2^{20}

                 = 2^1 \times 2^{20}\\\\= 2^{21}

The memory position for '2^n' could be 'n' m bits'  

It can use 2^{21} bits to address the memory location of 21.  

That is to say, the 2-mega-location memory needs '21' bits.  

Memory Length = 21 bit Address

In point c:

i^{th} element array addresses are given by:

\to address [i] = B+w \times (i-LB)

_{where}, \\\\B = \text {Base  address}\\w= \text{size of the element}\\L B = \text{lower array bound}

\to B=\$ 52\\\to w= 4 byte\\ \to L B= 0\\\to address  = 10

\to address  [10] = \$ 52 + 4 \times (10-0)\\

                       =   \$ 52   + 40 \ bytes\\

1 term is 4 bytes in 'MIPS,' that is:

= \$ 52  + 10 \ words\\\\ = \$ 512

In point d:

\to  base \ address = \$ t 5

When MIPS is 1 word which equals to 32 bit :

In Unicode, its value is = 2 byte

In ASCII code its value is = 1 byte

both sizes are  < 4 byte

Calculating address:

\to address  [5] = \$ t5 + 4 \times (5-0)\\

                     = \$ t5 + 4 \times 5\\ \\ = \$ t5 + 20 \\\\= \$ t5 + 20  \ bytes  \\\\= \$ t5 + 5 \ words  \\\\= \$ t 10  \ words  \\\\

3 0
3 years ago
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