How did the colonists feel about the powers of government and taxation?

Only the colonists' elected representatives should have the power to charge taxes.

They organized protests to speak out against unfair treatment.

They conducted the Boston Tea Party to damage British materials and pride.

Select all the ways The Sons of Liberty fought against the British (pick three)

They would intimidate tax collectors by burning effigies and threatening tar and feathers.

Who made up the majority of the Committee of 99?

Plantation owners and merchants

The government getting its power from the people is expressed in which important document?

The Declaration of Independence

Why were governments formed according to the Declaration of Independence?

protect a person's life, their freedom, and the right to find happiness

King George was the intended audience for which document...

The Declaration of Independence

They were all upper-class, white males.

They were all plantation owners.

Select all the true statements about South Carolina's signers of the Declaration of Independence. (pick three)

They all came from the Low Country.

Unit Three Set Two Review

Unit Two Set Three Review

People of Southeast Asia By the late 20th century, Southeast Asia’s population (including Indonesia and the Philippines) was approaching a half billion, or about one-twelfth of the world’s total. This population, however, was unevenly distributed within the region. By far the nation with the largest population was Indonesia, with about two-fifths of the regional total; in contrast, Brunei’s population was only a tiny fraction of that. Nearly half of the regional population was accounted for by the mainland states, with Vietnam and Thailand being the most populous. Settlement patterns Southeast Asia is predominantly rural: three-fourths of the people live in nonurban areas. Moreover, population is heavily clustered in fertile river valleys and especially in delta areas, such as those of the Mekong and Irrawaddy rivers. Historical, cultural, and environmental influences also have affected the settlement patterns. Java and other core areas such as the Bangkok (Thailand), Hanoi, and Manila metropolitan areas contain high population densities. While the rate of urbanization in Southeast Asia is relatively low compared with those of other developing regions, it is increasing rapidly. Singapore is unique in that it is essentially totally urban. In addition, the Philippines has a much higher than average level of urbanization, in part because of its Spanish and American colonial history. The largest cities—Jakarta (Indonesia), Bangkok, and Manila—are among the world’s most populous. The growth of cities of all sizes is being fueled primarily by natural increase, but rural-urban migration also is a significant contributor. Rural dwellers continue to be attracted by the promise of employment and other opportunities, but for many migrants the informal (undocumented) economic sector in these large cities is the only hope for some form of employment. Settlement patterns in rural areas tend to be associated with agricultural practices. Shifting cultivation is still common in some parts of the region (notably the remote interior areas of Myanmar, Vietnam, and the island of Borneo), although the amount of land so utilized is gradually shrinking. The village is the unit of settlement and often functions collectively, and typically it is moved from time to time. By contrast, wet-rice cultivation, the dominant form of agriculture in Southeast Asia, is sedentary and results in relatively large rural agglomerations with well-developed village life and customs. Dry and upland farming often produces scattered homesteads. Population resettlement to provide agricultural employment and access to land is important in some Southeast Asian countries, notably Indonesia, Malaysia, and Vietnam. By far the largest program has been conducted in Indonesia, where more than four million people have been voluntarily resettled from Java and Bali to the less populated islands. Despite considerable success, the program has been plagued by such problems as improper site selection, environmental deterioration, migrant adjustment, land conflicts, and inadequate financing. A program in Malaysia also has been quite successful, in part because it has set much smaller resettlement targets and has been better funded. Vietnamese development policy also has utilized the resettlement of people in an effort to revitalize areas outside the major population centres.

In this video we take a look at the 0:02 fetch to code 0:03 execute cycle including its effect on 0:06 the various registers we've previously 0:12 [Music] 0:14 discussed a computer is defined Definition 0:17 as an electronic device that takes an 0:20 input 0:22 processes data 0:25 and delivers output 0:29 in this simple example you can see we're 0:31 taking the input 5 0:35 we're multiplying it by 2 that's our 0:37 process 0:39 and we're outputting 10. 0:44 but this could be way more complex for 0:46 example of a game console 0:48 the input could be the buttons you press 0:50 on a controller 0:53 the processes would then be carried out 0:55 by the console itself 0:59 and the output would be some form of 1:01 update to a monitor 1:02 and sound out for a speaker possibly 1:04 vibration feedback through the 1:06 controller 1:10 to process data a computer follows a set 1:13 of instructions 1:14 known as a computer program 1:18 if we take the lid off a typical desktop 1:20 computer we can identify 1:22 two critical components the memory 1:26 that stores the program and the central 1:29 processing unit or processor 1:31 which is under this large fan and 1:33 carries out the instructions 1:37 a computer carries out its function by 1:40 fetching 1:41 instructions decoding them and then 1:43 executing them 1:44 in a continuous repetitive cycle 1:46 billions of times a second 1:48 let's look at each of these stages in a 1:50 little more detail Fetch 1:53 so let's start with the fetch stage the 1:55 very first thing that happens 1:57 is the program counter is checked as it 2:00 holds the address 2:01 of the next instruction to be executed 2:07 the address stored is then copied into 2:09 the memory address register 2:14 the address is then sent along the 2:16 address bus to main memory 2:18 where it waits to receive a signal from 2:21 the control 2:22 bus so it knows what to do 2:27 as we want to read the data that's 2:29 stored in memory address 2:30 0 0 0 0 the control unit sends 2:34 a read signal along the control bus to 2:36 main memory 2:41 now main memory knows the data needs to 2:44 be read 2:45 the content stored in memory address 000 2:49 can be sent along the data bus to the 2:51 memory data register 2:56 now as we're currently in the process of 2:58 fetching an instruction 3:00 the data received by the memory data 3:03 register gets copied 3:04 into the current instruction register 3:11 the instruction effectively has now been 3:14 fetched from memory 3:16 just before we proceed to the decode 3:18 phase we now 3:19 increment the program counter so that 3:22 the address it contains 3:24 points to the address of the next 3:26 instruction which will need to be 3:30 executed 3:32 the instruction now being held in the 3:33 current instruction register 3:35 is ready to be decoded 3:39 now as we mentioned in the previous 3:41 video the instruction is made up of two 3:43 parts 3:44 we have the op code that's what it is we 3:47 need to do 3:50 and we have the operand what are we 3:53 going to do it to 3:55 now the operand could contain the actual 3:57 data 3:58 or indeed it could contain an address of 4:01 where the data is to be found 4:06 by decoding this instruction we can see 4:08 the operation we need 4:10 is a load operation so we need to load 4:14 the contents of memory location0101 4:18 into the cpus accumulator 4:25 in the exam a simple model will be used 4:27 to describe the 4:29 structure of any given instruction 4:32 you're not going to be expected to 4:34 define how an opcode is made up 4:36 but simply to interpret opcodes in the 4:39 given context of an exam 4:40 question in the example here 4:44 you can see there's a total of 16 4:46 different opcodes available 4:48 and this is because we're using four 4:50 bits for our representation 4:56 so now we've fetched the instruction and 4:59 we've decoded it so we know what we need 5:00 to do 5:01 we're finally ready to execute it 5:05 so we now send address 0101 5:08 to the memory dress register 5:13 now we're in the memory address register 5:15 we can finally send the address 5:18 down the address bus to main memory 5:24 this time we want to read the data 5:26 that's stored in memory 5:28 and so the control unit again sends a 5:30 read signal along the control bus 5:36 so main memories now receive an address 5:38 and a read signal 5:40 so the content stored at memory location 5:43 0101 5:44 can now be sent along the data bus back 5:46 to the cpu 5:47 and into the memory data register 5:54 finally the contents of the memory data 5:56 register are copied to the accumulator 5:59 and this is one of a number of general 6:00 purpose registers found in the cpu 6:04 this first instruction is now complete Branching 6:11 so what does this program actually do 6:14 you should be able to work it through 6:16 carefully and figure it out 6:19 we're now pointing instructions zero 6:21 zero zero one in the program counter 6:23 and we're ready to fetch the second 6:25 instruction 6:27 at the end of this video we're gonna 6:29 provide you with the answer 6:34 so let's talk a second about programs 6:37 that branch 6:40 on the left here we have a very simple 6:42 piece of pseudo code 6:44 line zero says first execute this line 6:46 of code 6:47 line 1 now execute this line and then 6:50 line 2 says 6:52 if the age is greater than 18 then 6:56 we're going to execute lines 3 and 4 6:58 otherwise 6:59 we're going to execute lines six and 7:02 seven 7:03 so this program doesn't necessarily 7:05 follow strictly in sequence from line 7:07 zero through to seven there's a chance 7:10 here the program may branch and jump 7:14 around 7:16 so we're going to pretend that this 7:17 program has been loaded into memory 7:20 each line of code on the left here has 7:23 ended up 7:24 as a location in memory now this is not 7:27 strictly how this would happen in this 7:28 one-to-one way 7:29 but for the purpose of example it's 7:31 absolutely fine 7:35 so the program counter starts by 7:37 pointing to memory address zero 7:39 and we fetch the first instruction 7:41 decode it and execute it 7:44 it then updates and tells us the next 7:47 instruction 7:48 is zero zero zero one because remember 7:50 the program counter is being incremented 7:52 so we fetch it decode it and we execute 7:55 line one of our program 7:59 we then fetch line two which in binary 8:01 is one 8:02 zero 8:06 now at this point depending on what 8:10 happens during the execution 8:11 of line two the program may be required 8:15 to fetch line three from memory or 8:18 line five from memory 8:25 so let's look at how this actually works 8:27 because we've said the program counter 8:28 simply gets incremented 8:31 well in the current instruction register 8:33 we have an instruction with the op code 8:36 0 1 1 0. 8:41 now when we look this up in the decode 8:43 unit we discover that this 8:45 code means branch always 8:51 this replaces the value held in the 8:54 program counter 8:56 with the contents of the operand that's 8:58 the second part of the instruction 9:01 from the current instruction register so 9:03 this case 9:04 one zero zero one 9:09 now when the next fetch cycle begins the 9:12 program counter is obviously checked 9:14 and as its contents have been previously 9:16 updated to a new memory location 9:19 and not simply incremented the program 9:22 effectively is able to jump 9:24 around memory 9:28 so having watched this video you should 9:30 be able to answer the following key 9:32 question 9:33 how does a cpu work 9:39 okay so let's um answer the question we 9:41 posed 9:42 earlier what did that program actually 9:48 do 9:50 so this is the first fetch to code 9:53 execute cycle 9:55 and this is the one that we ran through 9:57 in detail earlier 9:58 it effectively loaded the contents of 10:01 the memory 10:02 stored at location location0101 10:05 into the accumulator in other words 10:08 the dna number 3 is moved 10:11 from memory into the cpu 10:18 we then proceed onto the second fetch 10:20 decode execute cycle 10:23 now this one adds the contents of memory 10:27 located at 0 1 1 0 10:30 to the current contents of the 10:32 accumulator 10:34 so in other words the dna number one 10:38 because that's what's stored at address 10:40 zero one one zero 10:43 is added to the number three that was in 10:45 the accumulator 10:46 the results are stored back over the 10:48 accumulator 10:49 so effectively we've done three plus one 10:53 equals four 10:58 the third fetch to code execute cycle 11:00 stores the contents which are in the 11:02 accumulator 11:03 into memory location zero one one one 11:07 and that's because the op code the first 11:09 part of this current instruction 11:10 zero zero one one is the command to 11:13 store when we look it up in the decoder 11:15 unit 11:16 so in other words the result of the 11:17 previous calculation three plus one 11:19 equals four 11:20 is now written back into main memory 11:28 the fourth fetch decode execute cycle 11:30 outputs the contents of the accumulator 11:33 remember they were copied into main 11:34 memory but they're still held in the 11:35 accumulator 11:37 so in this simple abstraction the number 11:40 four is now 11:41 output to the user so they can see the 11:43 result of the calculation 11:49 the fifth and final fetch code execute 11:51 cycle 11:52 brings a halt to the current program 11:58 so this very simple program which has 12:01 five 12:02 fetch decode execute cycles has 12:04 performed the calculation 12:06 three plus one is then stored the result 12:09 in main memory 12:10 and displayed the result four to the 12:12 user 12:13 and in a high-level language this may 12:15 look something very similar to the 12:17 following two lines of code 12:20 sum variable equals num1 plus num2 12:24 print sum to the user 12:27 so you can start to get an appreciation 12:29 here of how the high level code you 12:32 write actually ends up being fetched 12:34 decoded 12:35 and executed inside a processor 12:38 of course your processor is doing 12:40 billions and billions of these 12:42 operations a second 12:43 which when you think about it is really 12:45 very impressive 12:52 [Music] 13:03 you. make 10 questions for a standerd of a level

Can you create an evaluation using this information PHONETICS VS. PHONOLOGY Whereas phonetics is the study of sounds that occur in language, phonology is the study of how these sounds are organized and how they function in language. It uses the classifications of sounds derived from phonetics to describe and analyze how sounds occur in speech. STRUCTURALIST PHONEMICS STRUCTURALIST PHONEMICS As linguists began to study sounds in fine detail, they recognized increasingly complex aspects of phonetic organization. For example, the sound /p/ appears in different varieties in English. STRUCTURALIST PHONEMICS One of the varieties of /p/ is indicated by [ph]. This sound is produced with an accompanying puff of air called aspiration, as in the words “pill,” and “peace.” Another sound, indicated by [p•], is produced when there is little or no aspiration; this sound occurs in a word like “spill.” A third major variety for the /p/ sound is the unreleased [p– ], which may occur at the end of a word like “stop.” To deal with these variations for the /p/ sound, the structuralists suggested the existence of an abstract unit which they termed a phoneme. STRUCTURALIST PHONEMICS A phoneme was defined by the structuralists as an abstract phonological unit that represents a class of real sounds, termed the allophones of a phoneme. The phoneme /p/ in English, then, is represented by the allophones [ph], [p•], and [p– ]. STRUCTURALISTS: MINIMAL PAIRS How do we know what these abstract units of sound called phonemes are? In order to find the phonemes of a language, the structuralists developed the concept of the minimal pair, defined as any two words that: a) Contain the same number of segments b) Differ in meaning c) Exhibit only one phonetic difference. STRUCTURALISTS: MINIMAL PAIRS In practical terms, phonemes distinguish meanings; and a phoneme can also be defined as the smallest meaning-distinguishing unit of sound. For instance, the words “pin” /pɪn/ and “bin” /bɪn/ mean different things, and the only one difference in these words occurs in the initial sounds. STRUCTURALISTS: MINIMAL PAIRS By using the concept of a minimal pair, we can determine that the three variations of the /p/ sound do not represent three phonemes. Certainly, it is possible to pronounce the word cap with either an aspirated [ph ] or unreleased [p– ]; however, the two forms [kæph ] and [kæp– ] are not a minimal pair, even though they involve different sounds, because they are identical in meaning. STRUCTURALISTS: FREE VARIATION The two forms [kæph ] and [kæp– ] are, therefore, said to exhibit free variation: that is, the pronunciation may vary without signifying a change in meaning. In other words, we may conclude that the unreleased [p– ] and the aspirated [ph ] are not representations of different phonemes in English; they are, in fact, allophones of one phoneme, /p/. STRUCTURALISTS: COMPLEMENTARY DISTRIBUTION When phonemes have more than one allophone in a language, the allophones are said to be in complementary distribution. Complementary distribution means that the allophones of a phoneme occur in different phonetic environments (that is, with different sounds surrounding them). TRANSFORMATIONAL- GENERATIVE PHONOLOGY TRANSFORMATIONAL-GENERATIVE PHONOLOGY Transformational-generative phonology is a relatively recent development in linguistic theory. Chomsky launched Transformational-Generative Grammar in 1957, but the earliest studies within this framework were largely concerned with syntax. A decade later, the first comprehensive transformational-generative treatment of English phonology appeared: Chomsky and Halle’s The Sound Pattern of English (1968). TRANSFORMATIONAL-GENERATIVE PHONOLOGY Transformational-generative phonologists strongly oppose the structuralists’ phonemic level. They replace this level by a series of rules that directly relate underlying representations to observed phonetic representations. The central mechanisms in transformational-generative phonology, then, are underlying representations and phonological rules. PHONOLOGICAL RULES A rule is an operational statement in which some linguistic entity is modified, resulting in a new linguistic entity. Rules may add elements, remove elements, or change elements. By using phonological rules, linguists attempt to demonstrate that there is order in linguistic phenomena and that linguistic patterns are systematic. PHONOLOGICAL DERIVATION A phonological derivation is an operation that begins with an underlying representation and, through the application of a set of specific rules, yields the actual sound the speaker produces. The representation of a phonological rule has the following general appearance. /A/ → [B] / C “A” changes to “B” under condition “C” PHONOLOGICAL RULE – EXAMPLE In most Southern dialects, the word ten is pronounced like the word tin. This is not an isolated fact, for den is pronounced like din and Ben is pronounced like bin, and so on. This very general fact can be represented by the phonological rule: /ɛ/ → [I] / ___ [n] den /dɛn/ → /dIn/ Ben /bɛn/ → /bIn/ ten /tɛn/ → /tIn/ /ɛ/ → [I] / ___ [n] - high - low - tense + front + high - tense + front + sonorant + anterior + coronal - continuant NOTATIONAL DEVICES IN PHONOLOGICAL RULES The statement of phonological rules can be complex, and linguists have developed several notational devices for writing them. Often, the following symbols will be necessary for stating the conditions under which rules apply: # indicates a word boundary + indicates an intraword boundary $ indicates a syllable boundary UNDERLYING REPRESENTATIONS AND RELATED ISSUES The transformational-generative description of phonology relates underlying representations to phonetic representations by rules. This can be represented in a simple example: In English, there are certain pairs of words like sign / signature, and malign / malignant that exhibit a regular alternation in their phonetic representations: [g] is present in the second member of the pairs but absent in the first member. UNDERLYING REPRESENTATIONS AND RELATED ISSUES To explain the relatedness of words such as sign / signature, we could claim that the underlying representation of the segment in all such pairs is /g/ and that a rule operates to delete /g/ before syllable-final nasals. Thus, the rule “/g/ is deleted before syllable-final nasal” would appear formally as: + voice - anterior →∅ ____ [+ nasal] $ - coronal UNDERLYING REPRESENTATIONS AND RELATED ISSUES On the left-hand side of the arrow, we place the features needed to uniquely specify /g/ among the consonants; that is, no other consonant has the features [+ voice], [- anterior], and [- coronal]. The symbols → mean that the sound /g/ changes to nothing or more properly “/g/ is deleted.” The horizontal line following the slash mark refers to the position of /g/ - namely, before a segment that is [+nasal]. Finally, this [+nasal] segment occurs before a syllable boundary, as indicated by $. A less formal way of writing this rule would be: /g/ → / _ [+nasal] $ Notice that this rule also helps describe such alternations as phlegm/phlegmatic and paradigm/paradigmatic. Application Activity: Think of other words in which this rule can be applied. Write the sound segments to prove /g/ is deleted. Another example is the process through which the prefix meaning “not” is added to words. This prefix alternates among the forms /Im/, /In/, and /Iŋ/, depending on the point of articulation of the initial segment of the following word. -If the segment begins in the extreme front part of the mouth (labials), the form is /Im/, as in improper. -If the segment begins in the extreme back part of the mouth (velars), the form is /Iŋ/, as in incomplete. -If the segment begins in the mid-region of the mouth (all other sounds), the form is /In/, as in indecent. *Exceptions:Words beginning with /r/ or /l/. Analyze the Word “in + complete,” for example. /n/ → [ŋ] / __ [k] - continuant - continuant - continuant + sonorant → + sonorant - sonorant + anterior - anterior - strident + coronal - coronal - coronal + tense THE VELAR SOFTENING RULE Still another example of alternation in English is found in pairs of words like “electric / electricity,” in which the segments /k/ and /s/ alternate. /k/ changes to [s] only before non- low, front vowels. THE VELAR SOFTENING RULE /k/ → [s] / __ - continuant + continuant - strident → - sonorant V - anterior + anterior - low - coronal + coronal - back

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