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3 years ago

Which letters name the drive where windows most commonly stores data and programs

Computers and Technology

2 answers:

ICE Princess25 [194]3 years ago

6 0

Hi,

Answer: C.

My work: There were two drives by the names of A and B, but later they created Drive C.

I Hope I Helped!

lubasha [3.4K]3 years ago

5 0

Hello!

I believe the correct answer is C.

I hope it helps!

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1. How many bits would you need to address a 2M × 32 memory if:

Dominik [7]

Answer:

a) 23 b) 21
a) 43 b) 42
a) 0 b) 0

Explanation:

1) How many bits is needed to address a 2M * 32 memory

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8; L = ( 2M * 32 ) / 8 = 2M * 4

hence number of bits = log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

2) How many bits are required to address a 4M × 16 main memory

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

3) How many bits are required to address a 1M * 8 main memory

1M = 1^1 * 1^20 , item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0

3 years ago

How do you know when a spreadsheet object is active in a powerpoint presentation?

Ira Lisetskai [31]

It should be under lined, when you press it it should have a different colour

4 0

4 years ago

Using Amdahl’s Law, calculate the speedup gain of an application that has a 60 percent parallel component for (a) two processing

Nina [5.8K]

Answer:

a) Speedup gain is 1.428 times.

b) Speedup gain is 1.81 times.

Explanation:

in order to calculate the speedup again of an application that has a 60 percent parallel component using Anklahls Law is speedup which state that:

$t=\frac{1 }{(S + (1- S)/N) }$

Where S is the portion of the application that must be performed serially, and N is the number of processing cores.

(a) For N = 2 processing cores, and a 60%, then S = 40% or 0.4

Thus, the speedup is:

$t = \frac{1}{(0.4 + (1-0.4)/2)} =1428671$

Speedup gain is 1.428 times.

(b) For N = 4 processing cores and a 60%, then S = 40% or 0.4

Thus, the speedup is:

$t=\frac{1}{(0.4 + (1-0.4)/4)} = 1.8181$

Speedup gain is 1.81 times.

8 0

4 years ago

What did the strict study generally find about the effect of internet use on sleep?

rjkz [21]

A. should be the answer hope this helps :)

4 0

4 years ago

Read 2 more answers

An engineer is designing an HTML page and wants to specify a title for the browser tab to display, along with some meta data on

Nady [450]

An HTML is made up of several individual tags and elements such as the head, body. form, frame and many more.

In an HTML page, the meta element and the title element are placed in the head element.

An illustration is as follows:

< head >

< title > My Title < /title >

< meta charset="UTF-8" >

< / head >

The head element contains quite a number of elements and tags; some of them are:

meta
title
style
script
base

And so on.

Hence, in order to use a meta-data element, the meta element has to be placed within the head element.