399 Words2 Pages

Exercise 1.3.1
What is the decimal value of Byte 1 by itself? What is the decimal value of Byte 2 by itself? -6400 -233
Exercise 1.3.2 What is the decimal equivalent of the binary sequence in Figure 1- 12 (the combined sequence of Byte 1 and Byte 2 as a single decimal value)? How does this compare to the individual values of Byte 1 and Byte 2? -6633 -There is an increase of bits. From 8 to 16, which increase the decimal number.
Exercise 1.3.3
Given a device with a storage capacity of 120 MB, how many bytes can be stored on this device? Show your calculations. -1024KB*120=122880 122880KB*1024=125829120B
Exercise 1.3.4
Given a computer with a disk capacity of 16 GB and a word size of 32 bits, how many words can be stored on the disk? Show your calculations. -16*1024MB=16384MB -16384MB*1024KB=16777216KB -16777216KB*1024B=1.71798918E10 -1.7179891E10/32=536870912 words
Exercise 1.3.5
Represent the binary value 110110 2 in hexadecimal. Show the steps of conversion that you used. -0011 0110 0011 represents 3 0110 represents 6
- 36 Exercise 1.3.6
Represent the hexadecimal value f6 16 in binary and decimal. Show the steps of conversion that you used.
- The best way I feel to do this propose is to change it to binary first.
- F is 1111
- 6 is 0110
- 11110110 in binary
- Then do the decimal step , 246
Lab 1.3 Reviews
Explain why it is important to know how many system words will fit in a primary storage device on a computer (such as the hard drive). -So that you know how much ad primary storage unit can hold.
Explain why more information can be contained in multiple bytes joined together than in a single byte. -Multiple bytes has more ones and zero as bytes with holds more information a single byte only has a series of 8 bit which has less data then a series of multiple sets of 8bit.
Explain why it is more

Related

## Nt1210 Unit 1 Rfeview Questions

1378 Words | 6 Pagesa. 8 bits per double word b. 32 bits per word c. 64 bits per quadruple word d. 4 bits per byte 4. Which of the following answers are true about random-access memory (RAM) as it is normally used inside a personal computer? (Choose two answers.)

## Nt1210 Unit 1 Assignment

902 Words | 4 PagesAnswer These Questions 1. Which of the following is true about 1 bit? Represents one binary digit 2. Which of the following terms means approximately 106 bytes Megabyte 3. Which answer lists the correct number of bits associated with each term?

## Nt1210 Unit 1 Assignment 1

320 Words | 2 Pages20. A - Information that identifies each individual pixel on the computer display B - A binary code for each pixel, defining its color RAM = Random Access Memory OS = Operating System CPU = Central Processing Unit KB = Kilobyte MB = Megabyte GB = Gigabyte TB = Terabyte USB = Universal System Bus HDD = Hard Disk Drive ODD = Optical Disk Drive I/O = Input/Output DPI = Dots per

## It/240 Tcp/Ip Lan Plan

255 Words | 2 PagesCheckpoint TCP/IP LAN Plan 1. An IPv6 address is made up of how many bits? c. 64. It has four octets with 16 bits in each one. 2.

## Lab 1.1 Essay

280 Words | 2 PagesRandy Michael NT 1210 Lab 1.1 Professor Chibuzo Onukwufor 4/1/15 Lab 1.1 1: Convert the decimal value 127 to binary. Explain the process of conversion that you used. Decimal Number | Binary Number | Remainder | 127 - | 64 | 63 | 63 - | 32 | 31 | 31 - | 16 | 15 | 15 - | 8 | 7 | 7 - | 4 | 3 | 3 - | 2 | 1 | 1 - | 1 | 0 | Binary | 27 | 26 | 25 | 24 | 23 | 22 | 21 | 20 | Decimal | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | Conversion | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | I took the decimal and divided it by two giving 1 for the remainders and 0 if it did not have a remainder. 2: Explain why the values 102 and 00102 are equivalent. They are equivalent because they represent the powers of 10 3: Based on the breakdown of the decimal and binary systems in this lab, describe the available digit values and the first four digits of a base 5 numbering system.

## Nt1230 Unit 3 Assignment 1 Ipv6 Addressing

305 Words | 2 Pages2. Approximately how many IPv4 addresses are possible? 32 bit. About 4.294 Billion addresses. 3.

## Nt1210 Unit 3 - Assignment 1: Ipv6 Addressing

371 Words | 2 PagesThe IPv4 address space is 32 bits. 2^32 = 4,294,967,296 3. Approximately how many IPv6 addresses are possible? The IPv6 address space is 128 bits. 2^128 = 340,282,366,920,938,463,463,374,607,431,768,211,456 4.

## Nt1210 Lab 1.1

695 Words | 3 Pages3. Based on the breakdown of the decimal and binary systems in this lab, describe the available digit values and the first four digits of a base 5 numbering system. You can use the binary system as a reference, where the available digit values are 0 and 1 and the first four digits are 1, 2, 4, and 8? = 4. Using the Internet and the Help files in Excel, explain why creating a converter from decimal to binary would be more difficult to construct.

## It104 Wk 4 Hw

432 Words | 2 Pages6. What will the following pseudocode program display? Module Main() Declare Integer x = 1 Declare real y = 3.4 Display x, “ ”, Y Call changeUs (x, y) Display x, “ ”, y End Module Module changeUs(Integer a, Real b) Set a = 0 Set b = 0 Display a, “ “, b End Module 7. What will the following psudocode program display? Module main() Declare Integer x = 1

## 60-646 Lesson 6

855 Words | 4 PagesLESSON 6 Knowledge Assessment Matching: Complete the following exercise by matching the terms with their corresponding definitions. 1. __G___ switched fabric 2. __D___ VDS hardware provider 3. __F___ disk duplexing 4.

### Nt1210 Unit 1 Rfeview Questions

1378 Words | 6 Pages### Nt1210 Unit 1 Assignment

902 Words | 4 Pages### Nt1210 Unit 1 Assignment 1

320 Words | 2 Pages### It/240 Tcp/Ip Lan Plan

255 Words | 2 Pages### Lab 1.1 Essay

280 Words | 2 Pages### Nt1230 Unit 3 Assignment 1 Ipv6 Addressing

305 Words | 2 Pages### Nt1210 Unit 3 - Assignment 1: Ipv6 Addressing

371 Words | 2 Pages### Nt1210 Lab 1.1

695 Words | 3 Pages### It104 Wk 4 Hw

432 Words | 2 Pages### 60-646 Lesson 6

855 Words | 4 Pages