Promosi bisnis online pada scope openchat.

No Problem, Jika tema openchat berhubungan

Delete, Jika tema openchat berhubungan

Suspend, Jika tema openchat berhubungan

No Problem, Jika tema openchat tidak berhubungan

Fungsi eskalasi pada openchat?

Untuk konten yang bersifat ambigu dan high obscene

Untuk konten yang bersifat ambigu dan obscene

Untuk konten yang bersifat spamming

Untuk konten yang bersifat offensive

Contoh konten yang bukan termasuk kedalam kategori Assisted of Illegal Activities adalah?

Mengajak orang untuk bergabung menjadi member MLM

Penambahan restriction terbaru yang digunakan adalah?

Collecting personal information

Pada scope New Openchat, user dilarang?

Memasang foto pribadi sebagai profile image Openchat

Menuliskan nama keluarga besar sebagai tema Openchat

Menulis deskripsi dengan kombinasi symbol

Mengundang user lain untuk bergabung Openchat

Pada scope Main Report, konten yang harus di eskalasi adalah?

Kalimat yang panjang lebar, berulang-ulang dan tidak ada artinya (3 kali report)

Percakapan antara pelacur dengan calon pelanggan (3 kali report oleh pelacur)

Video peperangan dokumenter jaman dulu pada group chat (3 kali report)

Menyebarkan hoax pada group chat (3 kali report)

Apa yang harus dilakukan jika terjadi kendala koneksi pada saat tidak ada support ditempat?

Melaporkan detail waktu mulai kendala dan akibat yang timbul dari sistem pada group IT Issue

Meninggalkan pekerjaan untuk break makan siang

Tidak melakukan apapun, tetap bekerja walaupun ada kendala sistem

Melaporkan kendala kepada teman sebelah untuk kemudian kembali bekerja

LINE Plus Quiz! December 2019

LINE Plus Monitoring Quiz - November'19

Timmy: Did you finish your homework, Jimmy? Jimmy: Yes, I finished it. Timmy: Can you help me now? I have to do my math homework and then I can jump rope in the park. Jimmy: OK! What's the problem? Timmy: What is ten plus three? Jimmy: Here, look at the number line. Ten plus three equals thirteen. Timmy: Twenty minus four equals fifteen. Is that right? Jimmy: No, twenty minus four equals sixteen.

Policy for Bloodborne Pathogen Exposure Incident as per OSHA regulation (29 CFR 1910.1030) Purpose: To ensure that ASC staff members are protected against potential exposure to bloodborne pathogens per OSHA regulations (29 CFR 1910.1030). Scope: This policy applies to all ASC staff members who may be exposed to blood or other potentially infectious materials during their duties. Policy: An exposure incident is defined as a specific eye, mouth, other mucous membranes, non-intact skin, or parenteral contact with blood or other potentially infectious materials that results from the performance of an employee's duties. Any spill or accident that results in an exposure incident must be immediately reported to the Infection Control Nurse, first-line leader, or another responsible person. The employer shall make available the hepatitis B vaccine and vaccination series to all employees who have occupational exposure and post-exposure evaluation and follow-up to all employees who have had an exposure incident. The employer shall provide a confidential medical evaluation and follow-up for the exposed employee, which shall include at least the following elements: • Documentation of the route(s) of exposure and the circumstances under which the exposure incident occurred. • Identification and documentation of the source individual, unless the employer can establish that identification is infeasible or prohibited by state or local law. The source individual's blood shall be tested as soon as feasible and after consent is obtained to determine HBV and HIV infectivity. • Collection and testing of blood for HBV and HIV serological status. • If the employee consents to baseline blood collection but does not consent to HIV serologic testing, the sample shall be preserved for at least 90 days. If, within 90 days of the exposure incident, the employee elects to have the baseline sample tested, such testing shall be done as soon as feasible • Post-exposure prophylaxis, when medically indicated, as recommended by the U.S. Public Health Service. • Counseling. • Evaluation of reported illnesses. The employer shall ensure that the healthcare professional evaluating an employee after an exposure incident is provided with the following: A copy of OSHA regulation 1910.1030 A description of the exposed employee's duties as they relate to the exposure incident Documentation of the route(s) of exposure and circumstances under which exposure occurred Results of the source individual's blood testing, if available. All medical records are relevant to the appropriate treatment of the employee, including vaccination status, which is the employer's responsibility to maintain. The employer shall obtain and provide the employee with a copy of the evaluating healthcare professional's written opinion within 15 days of the completion of the evaluation. The healthcare professional's written opinion for Hepatitis B vaccination shall include the following: Whether it is indicated for the employee If the employee has received such a vaccination The healthcare professional's written opinion for post-exposure evaluation and follow-up shall include the following: That the employee has been informed of the results of the evaluation That the employee has been told about any medical conditions resulting from exposure to blood or other potentially infectious materials which require further evaluation or treatment All other findings or diagnoses shall remain confidential and not be included in the written report. An employer must establish and maintain accurate medical records for each employee with occupational exposure. Records should include the employee's Name, hepatitis B vaccination status and dates, results of medical testing and follow-up procedures, healthcare professional's written opinion, and information provided to the healthcare professional. Records must be kept confidential and not disclosed without the employee's written consent, except as required by law. Records must be kept for at least the duration of the employee's employment plus 30 years. Form 7.041 Employee Consent Form for Testing for HBV and HIV Serological Status Following Accidental Exposure I, __________________________, understand that I have been involved in an accidental exposure incident and may be at risk for contracting Hepatitis B Virus (HBV) and Human Immunodeficiency Virus (HIV) under 29 CFR 1910.1030. Therefore, following OSHA standards, I am being offered the opportunity to be tested for these viruses. I understand that the testing will involve a blood sample and that the results will be kept confidential and will only be shared with authorized personnel. I also understand that testing is voluntary and that I have the right to refuse to test. By signing this form, I consent to be tested for HBV and HIV following the accidental exposure incident. Signed: __________________________ Patient's Name: __________________________ Form 7.042 Patient Consent Form for Testing for HBV and HIV Serological Status Following Accidental Exposure I, __________________________, understand that a staff member involved in an accidental exposure incident may be at risk for contracting Hepatitis B Virus (HBV) and Human Immunodeficiency Virus (HIV) following 29 CFR 1910.1030. Therefore, by OSHA standards, the staff member may be offered the opportunity to be tested for these viruses. I also understand that testing of my blood is necessary to determine if I am infected with HBV and HIV. The results will be kept confidential and only shared with authorized personnel. I understand that testing is voluntary and that I have the right to refuse to test. By signing this form, I consent to the staff member being tested for HBV and HIV and to my blood testing following the accidental exposure incident. Signed: __________________________ Form7.043 Refusal of Testing Patient/Employee (Circle One) I,_____________________________________, understand that I have the right to refuse testing for Hepatitis B Virus (HBV) and Human Immunodeficiency Virus (HIV) following an accidental exposure incident per 29 CFR 1910.1030. I understand that if I refuse to test, it may impact my ability to receive appropriate medical treatment and the healthcare facility's power to respond to the exposure incident. Following 29 CFR 1910.1030, The source individual's blood shall be tested as soon as feasible and after consent is obtained to determine HBV and HIV infectivity. The employer shall establish that legally required consent cannot be obtained if permission is not obtained. When the source individual's consent is not required by law, the source individual's blood, if available, shall be tested, and the results documented. Signed: __________________________

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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