Which of the following is a consequence of frequent incumbent re-election in the House of Representatives?

Quality challengers choose not to run and wait until the seat is open.

Elections are completely predictable in the House of Representatives and never see a major shift.

The majority of incumbents fail to face a challenger due to an understanding that they will win.

Representatives in the House of Representatives don't feel the need to add pork to their legislation for their home districts.

In which of the following scenarios is the incumbency advantage likely to be most significant?

The Senator from a state with a primarily agricultural economy serves on the Senate Committee on Agriculture, Nutrition, and Forestry.

At the end of his second full term, the president's approval rating is almost 60 percent.

The current state legislator is involved in a highly publicized scandal involving misuse of public funds.

The challenger for a seat in the House of Representatives has lived in the legislative district for 30 years.

Which of the following describes a method by which state legislatures can affect presidential elections?

State legislatures can change electoral rules that may affect voter turnout in presidential elections.

State legislatures decide the number of electors that will represent their state in the electoral college.

State legislatures determine the date on which the general election for president is held.

State legislatures can limit the number of voters by increasing the voting age to 21.

Congressman Jefferson Smith is a Republican running for re-election in a state traditionally controlled by his own party to serve a fourth consecutive term in Congress. However, he is running in a presidential election year where a very popular Democratic president is likely to be re-elected. Which of the following statements describes a prime factor behind the incumbency advantage Congress Smith is likely to enjoy in his re-election bid?

Congressman Smith serves on some important congressional committees.

Congressman Smith can use the franking privilege to cover the cost of mailing campaign literature to his constituents.

Congressman Smith can "ride the coattails" of a popular incumbent president.

Congressman Smith has a lengthy voting record to boast about.

Which of the following best describes the impact of caucuses in the process of electing the president?

Because the first event of the presidential election season is a caucus, it can create significant momentum for the winning candidate from each party.

Because they typically generate a greater turnout than primaries, caucuses are more influential in determining each party's nominee.

The majority of delegates to each party's nominating convention comes from caucuses.

The relatively small number of caucuses that are held mean that they do not influence the choice of the presidential candidate.

Residents of a South-west state vote in the presidential election where the Democratic candidate receives 48% of the popular vote, the Republican candidate receives 46%, and the Libertarian candidate receives 6%. Which of the following correctly describes how the electoral votes would be allocated as a result?

The Democratic candidate will receive all of the states electoral votes.

The Democratic and Republican candidates will receive electoral votes proportionate to the percentage of the popular vote each received while the Libertarian candidate will receive no votes because she received less than 10% of the popular vote.

Each candidate will receive a number of the state's electoral votes proportionate to the percentage of the popular vote received.

The state will hold a runoff election to determine which candidate receives its electoral votes since no candidate received a majority.

Which of the following is a consequence of using a caucus as a primary?

They typically occur at a specific date and time which may deter some attendance.

Only those pre-registered to vote for a specific party are able to attend.

All candidates are place on the same ballot and the top-two vote recipients move on to the general election.

The winners tend to not win the general election.

A party’s congressional campaign committee typically does all of the following except

select the site of the party’s national convention

target potentially competitive seats for which party candidates may contend

raise funds on behalf of party candidates

allocate how much financial support a candidate will receive

Which of the following scenarios demonstrates the greatest advantage Congressional incumbents have seeking re-election?

Incumbents have better name recognition and can remind voters of the things they have accomplished while in Washington, D.C.

Incumbents receive public funds from the Federal Elections Commission which challengers are not eligible for.

Incumbents receive more funds from Political Action Committees than their challengers.

Super Pac spending always favors incumbents.

In a political climate marred by gridlock and partisanship, you are a political consultant employed by a Republican challenger who wants to unseat the eight-term incumbent Republican U.S. Representative in the next general election. Which campaign strategy below would be the advice for your candidate?

Present yourself as a moderate Republican who wants break the status quo and work with the Democrats to solve the problems of the nation and defeat the incumbent in the Republican primary or caucus of your state.

Run as a Libertarian touting your liberal economic platform of increasing taxes on the rich to fund social services.

Run as an Independent that wants to do what is best for America and work to end the gridlock and partisanship in Washington.

Attack the infrastructure spending the representative has brought to your state as typical pork-barrel legislation or the Washington establishment.

Unit Five: 5.9 Congressional elections (13)

1 What is the Australian Qualifications Framework (AQF)? 2 Describe three nationally accredited training products. 3 Define CBA. How is CBA different from other types of assessment? Briefly describe how the four (4) principles of assessment impact the CBA process. 4 Briefly describe how the rules of evidence impact the collection of evidence. 5 Describe different methods of gathering evidence for CBA . Describe the three main assessment methods. Describe the two different assessment pathways. What is recognition of prior learning (RPL), and what are the benefits? 6 What are the three endorsed components of training packages? 7 Briefly describe the components of a unit of competency (UofC). 8 What is the ACSF? Identify the five core skills. 9 How do the dimensions of competency guide assessment practices? 10 What is the purpose and context of assessment? How can you contextualise an assessment while maintaining the integrity of the Unit of Competency? 11 What is the difference between assessment tools and assessment instruments? What are the essential parts of an assessment tool? 13 What is the purpose of mapping against units of competency? 14 Describe reasonable adjustment in competency-based assessment. What are the limitations? 15 What would you do if a candidate provided you with evidence that was not authentic? 16 How does the marker’s guide support you in objective assessment judgement? 17 Give an example of inclusive language when giving assessment instructions. 18 What are the key WHS instructions that are required prior to commencing assessment? 19 What do you look for when clustering units of competency for an assessment tool? 20 What are the key characteristics of a cohort of learners that need to be considered in the planning an assessment tool?

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

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