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s344n2d4d5 [400]
3 years ago
12

Unscramble the words A: ESUOM RETUPOC B: KSID EVIRD

Computers and Technology
2 answers:
Mrrafil [7]3 years ago
5 0
A. Mouse coputer
B. Disk drive
kherson [118]3 years ago
3 0
A) Mouse copter
B) Disk Drive
hope this helps x
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Can anybody do Algorithm 2 for me (with Python).<br> Answer = 25 points.
Nostrana [21]

Answer:

age=int(input("Enter age"))

if age>=18:

    print("You are Young")

else

   print("You are child")

Explanation:

if you have any query or any problem kindly ask in comment

5 0
3 years ago
Write the class RoadSegmet. The class inherits from Transportation Class. The Class have a vector hourlySpeeds data field which
OLEGan [10]

Answer:

The length of time on the road segment is calculated by taking it's distance and dividing by the speed at the time of departure. This length of time is then added to the departureTime parameter to get the arrivalTime.

Define a RiverSegment class that inherits from the base TransportationClass from above. This class will have a vector<double> scheduledDepartureTimes data field that will keep track of the departure time (e.g. 10.5 represents the time 10:30) of all ferries on the river. You can assume these are sorted. Additionally, there is a _speed value that is the speed of the ferry at all times (ferries tend to be consistently slow). There is a setSpeed function to change the _speed value. Additionally, there is an addDepartureTime(double hour) function to add a departure time to the vector. Make sure the time are sorted

Computing the arrival time for the RiverSegment is a little more complicated since you need to wait for the next available departure time. Once you find the next available departure time (after the departure time passed in), you need to add the length of time on the river segment. This is done by dividing the _distance by the _speed. Add the time on the river to the departure time (from the vector of departure times) and this will give you the arrive time.

note: If there is no departure time scheduled for after your planned departure, you need to take the first available departure the following day. Our solution assumes the departure times are the same every day.

TransportationLink.h

#include <string>

using namespace std;

#ifndef __TRANSPORTATIONLINK_H__

#define __TRANSPORTATIONLINK_H__

const int HOURS_PER_DAY = 24;

const int MINS_PER_HOUR = 60;

const int MINS_PER_DAY = MINS_PER_HOUR * HOURS_PER_DAY; //24 * 60

class TransportationLink {

protected:

string _name;

double _distance;

public:

TransportationLink(const string &, double);

const string & getName() const;

double getDistance() const;

void setDistance(double);

// Passes in the departure time (as minute) and returns arrival time (as minute)

// For example:

// 8 am will be passed in as 480 (8 * 60)

// 2:30 pm will be passed in as 870 (14.5 * 60)

virtual unsigned computeArrivalTime(unsigned minute) const = 0;

};

5 0
2 years ago
Modify the list according to the following: append the smallest value at the end of the list when all numbers in the list are ne
ddd [48]

Answer:

The solution code is written in Python 3:

  1. def modifyList(listNumber):
  2.    posCount = 0
  3.    negCount = 0
  4.    for x in listNumber:
  5.        if x > 0:
  6.            posCount += 1
  7.        else:
  8.            negCount += 1
  9.    
  10.    if(posCount == len(listNumber)):
  11.        listNumber.append(max(listNumber))
  12.    
  13.    if(negCount == len(listNumber)):
  14.        listNumber.append(min(listNumber))
  15.    
  16.    print(listNumber)
  17. modifyList([-1,-99,-81])
  18. modifyList([1,99,8])
  19. modifyList([-1,99,-81])

Explanation:

The key step to solve this problem is to define two variables, posCount and negCount, to track the number of positive value and negative value from the input list (Line 2 - 3).

To track the posCount and negCount, we can traverse through the for-loop and create if else statement to check if the current number x is bigger than 0 then increment posCount by 1 otherwise increment negCount (Line 5- 9).

If all number in the list are positive, the posCount should be equal to the length of the input list and the same rule is applied to negCount. If one of them happens, the listNumber will append either the maximum number (Line 11 -12) or append the minimum number (Line 14-15).

If both posCount and negCount are not equal to the list length, the block of code Line 11 -15 will be skipped.

At last we can print the listNumber (Line 17).

If we test our function using the three sets of input list, we shall get the following results:

[-1, -99, -81, -99]

[1, 99, 8, 99]

[-1, 99, -81]

3 0
3 years ago
________ databases are better than relational databases at handling unstructured data such as audio clips, video clips, and pict
timofeeve [1]

Answer:

"Object-oriented" would be the correct choice.

Explanation:

  • An object-oriented database seems to be a database that subscribes to a framework containing object-depicted details. Throughout the context of the relational database management system, object-oriented is a unique product that is not as popular and most well-known as traditional web applications.
  • This indicates that internet connectivity to existing records has to implement the previously defined connections for interacting components established by that same containers.
3 0
3 years ago
Consider an error-free 64 kbps satellite channel used to send 512 byte data frames in one direction, with very short acknowledge
Vladimir [108]

Answer:

The answer is "2".

Explanation:

In the given question some information is missing, that is "The propagation time for satellite to earth" which is "270 milliseconds" so, the description to this question can be defined as follows:

Given values:

Bandwidth = 64 kbps

Data frames = 512 bytes  

Propagation Time (t​​​​​​p ) =270 ms

Change Bandwidth  kbps to bps:

1 kb= 1024 bytes

calculated bandwidth= 64 kbps = 64×1024 bps = 65536 bps

1 bytes = 8 bits

512 bytes = 512 × 8 =  4096 bits

Frame length = 4096 bits

Formula

Transmission time (T​​​​​​t) = Frame length /Bandwidth

Window size = 1+2a

where a = Propagation time/Transmission time

Calculate Transmission time:

Transmission time (T​​​​​​t) = 4096 / 65536

Transmission time (T​​​​​​t)= 625 m.sec

Calculate Window size:

Window size = 1+2(270/625)

Window size = 1+2(0.432)

Window size = 1+0.864

Window size = 1.864

Window size = 2

3 0
3 years ago
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