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fgiga [73]
2 years ago
6

Sarah is developing a Risk Assessment for her organization. She is asking each department head how long can they be without thei

r primary system. What recovery goal is Sarah trying to determine.
Engineering
1 answer:
Natali [406]2 years ago
8 0

Answer:

Sarah is asking each department head how long they can be without their primary system. Sarah is trying to determine the Recovery Time Objective (RTO) as this is the duration of time within which the primary system must be restored after the disruption.

Recovery Point Objective is basically to determine the age of restoration or recovery point.

Business recovery and technical recovery requirements are to assess the requirements to recover by Business or technically.

Hence, Recovery Time Objective (RTO) is the correct answer.

You might be interested in
Modify the Rainfall Statistics program you wrote for Programming Challenge 2 of Chapter 7 . The program should display a list of
rjkz [21]

Answer:

#include<iostream>

#include <iomanip>

using namespace std;

const int NUM_MONTHS = 12;

double getTotal(double [], int);

double getAverage(double [], int);

double getLargest(double [], int, int &);

double getSmallest(double [], int, int &);

double getTotal(int rainFall,double NUM_MONTHS[])

{

double total = 0;

for (int count = 0; count < NUM_MONTH; count++)

total += NUM_MONTH[count];

return total;

}

double getAverage(int rainFall,double NUM_MONTH[])

{getTotal(rainFall,NUM_MONTH)

average= total/NUM_MONTHS;

return average;

}

double getHighest(int rainFall, double NUM_MONTHS[]) //I left out the subScript peice as I was not sure how to procede with that;

{

double largest;

largest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month++ ){

                     if ( values[month] > largest ){

                 largest = values[month];

return largest;

          }

double getSmallest(int rainFall, double NUM_MONTHS[])

{

double smallest;

smallest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month){

                     if ( values[month] < smallest ){

                 smallest = values[month];

return smallest;

          }

 

int main()

{

double rainFall[NUM_MONTHS];

 for (int month = 0; month < NUM_MONTHS; month++)

  {

     cout << "Enter the rainfall (in inches) for month #";

     cout << (month + 1) << ": ";

     cin >> rainFall[month];

 

     while (rainFall[month] < 0)

     {

      cout << "Rainfall must be 0 or more.\n"

             << "Please re-enter: ";

      cin >> rainFall[month];

     }

  }

  cout << fixed << showpoint << setprecision(2) << endl;

  cout << "The total rainfall for the year is ";

  cout << getTotal(rainFall, NUM_MONTHS)

      << " inches." << endl;

   cout << "The average rainfall for the year is ";

  cout << getAverage(rainFall, NUM_MONTHS)

      << " inches." << endl;

   int subScript;

cout << "The largest amount of rainfall was ";

  cout << getLargest(rainFall, NUM_MONTHS, subScript)

      << " inches in month ";

  cout << (subScript + 1) << "." << endl;

  cout << "The smallest amount of rainfall was ";

  cout << getSmallest(rainFall, NUM_MONTHS, subScript)

      << " inches in month ";

  cout << (subScript + 1) << "." << endl << endl;

  return 0;

}

8 0
3 years ago
Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop pe
Sonja [21]

Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe\left ( D\right )=0.15m

Velocity of water in pipe\left ( V\right )=15cm/s

We know viscosity of water is\left (\mu\right )=8.90\times10^{-4}pa-s

Pressure drop is given by hagen poiseuille equation

\Delta P=\frac{128\mu \L Q}{\pi D^4}

We have asked pressure Drop per unit length i.e.

\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}

Substituting Values

\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}

\frac{\Delta P}{L}=0.1898 Pa/m

4 0
3 years ago
A 3.52 kg steel ball is tossed upward from a height of 6.93 meters above the floor with a vertical velocity of 2.99 m/s. What is
Dafna1 [17]

Answer : The final velocity of the ball is, 12.03 m/s

Explanation :

By the 3rd equation of motion,

v^2-u^2=2as

where,

s = distance covered by the object = 6.93 m

u = initial velocity  = 2.99 m/s

v = final velocity = ?

a = acceleration = 9.8m/s^2

Now put all the given values in the above equation, we get the final velocity of the ball.

v^2-(2.99m/s)^2=2\times (9.8m/s^2)\times (6.93m)

v=12.03m/s

Thus, the final velocity of the ball is, 12.03 m/s

7 0
3 years ago
Water (density p-1000 is discharging from through a hole at the bottom of a graduated 71 cylinder. The mass flow rate exiting th
alexira [117]

Answer:

Please see attachment

Explanation:

Please see attachment

4 0
2 years ago
What is 4 principles of experimental design
Mrrafil [7]

Answer:

manipulation, control , random assignment, and random selection

Explanation:

7 0
1 year ago
Read 2 more answers
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