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IrinaK [193]
3 years ago
10

Why did sugar planters come to Louisiana?​

Computers and Technology
2 answers:
Fiesta28 [93]3 years ago
5 0

Answer:

Because the soil there was rich and untouched.

Explanation:

Stella [2.4K]3 years ago
5 0
Yeah nobody had farmed there before so it was perfect for them to ues
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There will be 10 numbers stored contiguously in the computer at location x 7000 . Write a complete LC-3 program, starting at loc
Artist 52 [7]

Answer:

The LC-3 (Little Computer 3) is an ISA definition for a 16-bit computer. Its architecture includes physical memory mapped I/O via a keyboard and display; TRAPs to the operating system for handling service calls; conditional branches on N, Z, and P condition codes; a subroutine call/return mechanism; a minimal set of operation instructions (ADD, AND, and NOT); and various addressing modes for loads and stores (direct, indirect, Base+offset, PC-relative, and an immediate mode for loading effective addresses). Programs written in LC-3 assembler execute out of a 65536 word memory space. All references to memory, from loading instructions to loading and storing register values, pass through the get Mem Adr() function. The hardware/software function of Project 5 is to translate virtual addresses to physical addresses in a restricted memory space. The following is the default, pass-through, MMU code for all memory references by the LC-3 simulator.

unsigned short int get Mem Adr(int va, int rwFlg)

{

unsigned short int pa;

// Warning: Use of system calls that can cause context switches may result in address translation failure

// You should only need to use gittid() once which has already been called for you below. No other syscalls

// are necessary.

TCB* tcb = get TCB();

int task RPT = tcb [gettid()].RPT;

pa = va;

// turn off virtual addressing for system RAM

if (va < 0x3000) return &memory[va];

return &memory[pa];

} // end get MemAdr

Simple OS, Tasks, and the LC-3 Simulator

We introduce into our simple-os a new task that is an lc3 Task. An lc3 Task is a running LC-3 simulator that executes an LC-3 program loaded into the LC-3 memory. The memory for the LC-3 simulator, however, is a single global array. This single global array for memory means that alllc3 Tasks created by the shell use the same memory for their programs. As all LC-3 programs start at address 0x3000 in LC-3, each task overwrites another tasks LC-3 program when the scheduler swaps task. The LC-3 simulator (lc3 Task) invokes the SWAP command every several LC-3 instruction cycles. This swap invocation means the scheduler is going to be swapping LC-3 tasks before the tasks actually complete execution so over writing another LC-3 task's memory in the LC-3 simulator is not a good thing.

You are going to implement virtual memory for the LC-3 simulator so up to 32 LC-3 tasks can be active in the LC-3 simulator memory without corrupting each others data. To implement the virtual memory, we have routed all accesses to LC-3 memory through a get Mem Adr function that is the MMU for the LC-3 simulator. In essence, we now have a single LC-3 simulator with a single unified global memory array yet we provide multi-tasking in the simulator for up to 32 LC-3 programs running in their own private address space using virtual memory.

We are implementing a two level page table for the virtual memory in this programming task. A two level table relies on referring to two page tables both indexed by separate page numbers to complete an address translation from a virtual to a physical address. The first table is referred to as the root page table or RPT for short. The root page table is a fixed static table that always resides in memory. There is exactly one RPT per LC-3 task. Always.

The memory layout for the LC=3 simulator including the system (kernel) area that is always resident and non-paged (i.e., no virtual address translation).

The two figures try to illustrate the situation. The lower figure below demonstrates the use of the two level page table. The RPT resident in non-virtual memory is first referenced to get the address of the second level user page table or (UPT) for short. The right figure in purple and green illustrates the memory layout more precisely. Anything below the address 0x3000 is considered non-virtual. The address space is not paged. The memory in the region 0x2400 through 0x3000 is reserved for the RPTs for up to thirty-two LC-3 tasks. These tables are again always present in memory and are not paged. Accessing any RPT does not require any type of address translation.

The addresses that reside above 0x3000 require an address translation. The memory area is in the virtual address space of the program. This virtual address space means that a UPT belonging to any given task is accessed using a virtual address. You must use the RPT in the system memory to keep track of the correct physical address for the UPT location. Once you have the physical address of the UPT you can complete the address translation by finding the data frame and combining it with the page offset to arrive at your final absolute physical address.

A Two-level page table for virtual memory management.

x7000 123F x7000 0042

x7001 6534 x7001 6534

x7002 300F x7002 300F

x7003 4005 after the program is run, memory x7003 4005

x7004 3F19

7 0
3 years ago
Read 2 more answers
Windows hard disks can now use a variety of file systems, including FAT16, FAT32, ____, and Resilient File System.
marysya [2.9K]

Answer:

NTFS.

Explanation:

3 0
3 years ago
Recall that the name of a transport-layer packet is segment and that the name of a link-layer packet is frame. What is the name
AlladinOne [14]

Answer:

A network-layer packet is a data gram. A router forwards a packet based on the packet's IP (layer 3) address. A link-layer switch forwards a packet based on the packet's MAC (layer 2) address.

Explanation:

When the end system receives the unstructured bit stream from the physical wire, each layer removes the header information applicable to it until the application receives the data.

1. An application, such as an email program, creates data that will be sent by an end user, such as an email message. The Application layer places a header (encapsulation) field that contains information such as screen size and fonts, and passes the data to the Presentation layer.

2. The Presentation layer places layer 6 header information. For example, the text in the message might be converted to ASCII. The Presentation layer will then pass the new data to the Session layer (layer 5).

3. The Session layer follows the same process by adding layer 5 header information, such as information that the Session layer will manage the data flow, and passes this data to the Transport layer (layer 4).

4. The Transport layer places layer 4 information, such as an acknowledgement that the segment was received in the header, and passes it to the Network layer (layer 3).

5. The Network layer places layer 3 header information, such as the source and destination address so the Network layer can determine the best delivery path for the packets, and passes this data to the Data Link layer (layer 2).

6. The Data Link layer places layer 2 header and trailer information, such as a Frame Check Sequence (FCS) to ensure that the information is not corrupt, and passes this new data to the Physical layer (layer 1) for transmission across the media.

7. The bit stream is then transmitted as ones and zeros on the Physical layer. It is at this point that the Physical layer ensures bit synchronisation. Bit synchronisation will ensure the end user data is assembled in the correct order it was sent.

8. Steps 1 through 7 occur in reverse order on the destination device. Device B collects the raw bits from the physical wire and passes them up the

Data Link layer. The Data Link layer removes the headers and trailers and passes the remaining information to the Network layer and so forth until data is received by the Application layer. Eventually, Device B will receive an email notification displaying a message to indicate that a new email message has been received.

5 0
3 years ago
How many cost units are spent in the entire process of performing 40 consecutive append operations on an empty array which start
Alexus [3.1K]

Answer:

Explanation:

260 cost units, Big O(n) complexity for a push

6 0
3 years ago
Identify the variables listed below as either quantitative (discrete or continuous) or categorical (nominal or ordinal):
Akimi4 [234]

Answer:

Quantitative: Age, heart rate, number of pets, salary

 <em>discrete</em>: number of pets

 <em>continuous</em>: Age, heart rate, salary

Categorical: Gender, eye color, metal rankings

<em>Nominal:</em> Gender, eye color

<em>Ordinal:</em> metal rankings

Explanation:

Quantitative are those variables which are for quantity or number with units for measurement.

Categorical are qualitative variable which can be categorized or grouped in different groups or it may be ranking scales

5 0
3 years ago
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