What is a printed copy of computer output called

In C++ please (read the image below for instructions)

FromTheMoon [43]

Using the knowledge in computational language in C code it is possible to write a code that print the month name along with their rainfall inches. Then we have to calculate the total rainfall.

<h3>Writting the code:</h3>

#include <bits/stdc++.h>

using namespace std;

// function to print the rainfall of each month

void rainfallStatistics(string month[], double rainfall[]){

double sum = 0; // variable to store the total sum of rainfall

string maxMonth = month[0]; // variable to return month with maximum rainfall

double maxRainfall = INT_MIN; // variable to store the maximum rainfall

// running the loop till the last month

for(int i=0; i<12; i++){

// printing the month and rainfall

cout<< month[i] << " " << fixed << setprecision(2) << rainfall[i] << " inches" << endl;

// storing the sum

sum = sum + rainfall[i];

// checking the month with highest rainfall

if(rainfall[i] > maxRainfall){

maxRainfall = rainfall[i];

maxMonth = month[i];

}

// calculating the average of the rainfall

double average = sum / double(12);

cout << endl;

cout << "Total for the year " << sum << " inches" << endl;

cout << "Average rainfall per month " << average << " inches" << endl;

cout << "Month with the greatest rainfall was " << maxMonth << " with rainfall of " << maxRainfall << " inches" << endl;

}

// driver program

int main()

{

// array to store the nonth name

string month[12] = {"Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"};

// array to store the rainfall

double rainfall1[12] = { 4.50, 4.25, 3.26, 1.35, 0.80, 0.20, 0.10, 0.00, 0.40, 1.20, 2.96, 4.71 };

double rainfall2[12] = { 3.50, 1.25, 7.26, 8.35, 2.80, 0.00, 1.10, 4.00, 9.40, 2.20, 3.96, 4.71 };

cout << "Hayward Statistics" << endl;

cout << endl;

cout<< "Month" << " " << "Rainfall" << endl;

// calling the function for Hayward

rainfallStatistics(month, rainfall1);

cout << endl;

cout << "Houston Statistics" << endl;

cout << endl;

cout<< "Month" << " " << "Rainfall" << endl;

// calling the function for Houston

rainfallStatistics(month, rainfall2);

return 0;

}

See more about c code at brainly.com/question/19705654

#SPJ1

8 0

2 years ago

What is data anayltics

Lena [83]

Data analytics (DA) is the science of examining raw data with the purpose of drawing conclusions about that information. Data analytics is used in many industries to allow companies and organization to make better business decisions and in the sciences to verify or disprove existing models or theories.

7 0

3 years ago

Two machines can finish a job in StartFraction 20 Over 9 EndFraction hours. Working alone, one machine would take one hour long

Alborosie

Answer

5 hours

Explanation

The two working together can finish a job in

$\frac{20}{9} \: hours$

Also, working alone, one machine would take one hour longer than the other to complete the same job.

Let the slower machine working alone take x hours. Then the faster machine takes x-1 hours to complete the same task working alone.

Their combined rate in terms of x is

$\frac{1}{x} + \frac{1}{x - 1}$

This should be equal to 20/9 hours.

$\frac{1}{x} + \frac{1}{x - 1} = \frac{9}{20}$

Multiply through by;

$20x(x - 1) \times \frac{1}{x} +20x(x - 1) \times \frac{1}{x - 1} = 20x(x - 1) \times \frac{9}{0}$

$20(x - 1) +20x = 9x(x - 1)$

$20x - 20+20x = 9{x}^{2} - 9x$

$9{x}^{2} - 9x - 20x - 20x + 20= 0$

$9{x}^{2} - 49x + 20= 0$

Factor to get:

$(9x - 4)(x - 5) = 0$

$x = \frac{4}{9} \: or \: x = 5$

It is not feasible for the slower machine to complete the work alone in 4/9 hours if the two will finish in 20/9 hours.

Therefore the slower finish in 5 hours.

8 0

3 years ago

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

Switched backbone networks:_____.a. always use a ring topology.b. require much more management that do routed backbone networks.

Gre4nikov [31]

Answer:

c

Explanation:

spongo

7 0

3 years ago