Culver Inc., a greeting card company that follows ASPE, had the following statements prepared as at December 31, 2020:

CULVER INC.
Comparative Statement of Financial Position
December 31
2020      2019
Cash     $47,775         $25,090     
Accounts receivable   57,930         51,030     
Inventory      39,990         60,070     
Prepaid rent      5,210         4,100     
Equipment      161,990         130,120     
Accumulated depreciation–equipment      (35,210   )      (25,100   )
Goodwill      28,000         68,000     
    Total assets      $305,685         $313,310     
                        
Accounts payable      $46,190         $40,120     
Income tax payable      3,990         6,070     
Salaries and wages payable      8,090         4,090     
Short–term loans payable      7,930         10,030     
Long–term loans payable      68,000         87,000     
Common shares      130,000         130,000     
Retained earnings      41,485         36,000     
    Total liabilities and shareholders’ equity      $305,685         313,310     

CULVER INC.
Income Statement
Year Ending December 31, 2020
Sales revenue               $348,490
Cost of goods sold               165,000
Gross margin               183,490
Operating expenses               120,000
Operating income               63,490
Interest expense      $12,100           
Impairment loss–goodwill      40,000           
Gain on disposal of equipment      (2,400   )      49,700
Income before income tax               13,790
Income tax expense               4,105
Net income               $9,685

Additional information:

1.      Dividends on common shares in the amount of $4,200 were declared and paid during 2020.
2.      Depreciation expense is included in operating expenses, as is salaries and wages expense of $70,000.
3.      Equipment with a cost of $36,000 that was 70% depreciated was sold during 2020.

Prepare a statement of cash flows using the direct method. (Show amounts that decrease cash flow with either a - sign e.g. -10,000 or in parenthesis e.g. (10,000).)

Pro forma income statement   The marketing department of Metroline Manufacturing estimates that its sales in 2020 will be $1.53 million. Interest expense is expected to remain unchanged at $34,000, and the firm plans to pay $74,000 in cash dividends during 2020. Metroline Manufacturing's income statement for the year ended December31, 2019i is given (See belong Graph) ,along with a breakdown of the firm's cost of goods sold and operating expenses into their fixed and variable components. a. Use the percent-of-sales method to prepare a pro forma income statement for the year ended December 31, 2020 b. Use fixed and variable cost data to develop a pro forma income statement for the year ended December 31, 2020. c. Compare and contrast the statements developed in parts a. and b. Which statement probably provides the better estimate of 2020 income? Explain why.

Metroline Manufacturing Breakdown of Costs and Expenses into Fixed and Variable Components for the Year Ended December 31, 2019

Cost of goods sold:

Fixed cost $202,000

Variable cost 700000

Total cost $902,000

Operating expenses Fixed expenses $39,000

Variable expenses 80000

Total expenses $119,000

Metroline Manufacturing Income Statement for the Year Ended December 31, 2019

Sales revenue $1,396,000

Less: Cost of goods sold 902000

Gross profits $494,000

Less: Operating expenses 119000

Operating profits $375,000

Less: Interest expense 34000

Net profits before taxes $341,000

Less: Taxes (rate = 40%) 136400

Net profits after taxes $204,600

Less: Cash dividends 63000

To retained earnings $141,600

Mrs. Smith operates a business in a competitive market.  The current market price is $8.10.  At her profit-maximizing level of production, the average variable cost is $8.00, and the average total cost is $8.25.  Mrs. Smith should:                             

Many sports cars are convertibles. The air flow over such a car is significantly different depending on whether the convertible top is up or down. The engine of the 1000-kg car delivers 135 kW to the wheels, the car frontal area is 1.9 m² and rolling resistance is 2.5% of the car weight. The drag coefficient when the top is down is 0.43 and 0.31 when it is up. For 20 ºC air at one atmosphere, determine:

a. the maximum speed with the top up (in m/s)

b. the maximum speed with the top down (in m/s)

2.While stock selection is best approached from the bottom-up, ignoring the top-down can be extraordinarily expensive. The bottom-up can also inform the top-down. As Ben Graham pointed out “True bargain issues have repeatedly become scarce in bull markets (James Montier) 2.1. Differentiate between the bottom-up and top-down approach to fundamental analysis.(3) 2.2. Do you think investors should choose between the two approaches, or do you believe that they are complimentary? (3)

A car initially traveling at 40 m/s runs out of gas while traveling up a slope. It coasts for a distance of 500 m before it starts to roll back down.

a. What is the acceleration of the car?

b. What is the angle (incline) of the slope?

c. How long does it take after the car runs out of gas before it starts to roll back down?

d. Once the car starts rolling backwards, how long does it take before it passes its starting point, 500m down the slope?

A 50 kg crate slides down a ramp that is inclined by 30 degrees from the horizontal. The acceleration of the crate parallel to the surface of the ramp is 2.0m/s^2, and the length of the ramp is 10m.

Determine the kinetic energy accumulated by the crate when it reaches the bottom of the ramp if it started from rest at the top of the incline.

Calculate the amount of energy lost by the crate due to friction in its journey down the ramp.

Compute the magnitude of the frictional force that acts on the crate as it slides down the ramp and calculate the coefficient of kinetic friction between the ramp and the crate.

**java**

A bicycle combination lock has four rings with numbers 0 through 9. Given the actual numbers and the combination to unlock, print instructions to unlock the lock using the minimum number of twists. A “twist up” increases the number value of a ring, and a “twist down” decreases it. For example, if the actual number shown is 1729 and the desired combination is 5714, write your instructions like this:

Ring 1: Twist up 4 times

Ring 3: Twist down once

Ring 4: Twist up or down 5 times

Build a Date class and a main function to test it

Specifications
Below is the interface for the Date class: it is our "contract" with you: you have to implement everything it describes, and show us that it works with a test harness that puts it through its paces. The comments in the interface below should be sufficient for you to understand the project (use these comments in your Date declaration), without the need of any further documentation.

class Date
{
 private:
   unsigned day;
   unsigned month;
   string monthName;
   unsigned year;

 public:
   // creates the date January 1st, 2000.
   Date();


   /* parameterized constructor: month number, day, year 
       - e.g. (3, 1, 2010) will construct the date March 1st, 2010

       If any of the arguments are invalid (e.g. 15 for month or 32 for day)
       then the constructor will construct instead a valid Date as close
       as possible to the arguments provided - e.g. in above example,
       Date(15, 32, 2010), the Date would be corrected to Dec 31st, 2010.
       In case of such invalid input, the constructor will issue a console error message: 

       Invalid date values: Date corrected to 12/31/2010.
       (with a newline at the end).
   */
   Date(unsigned m, unsigned d, unsigned y);


   /* parameterized constructor: month name, day, year
 ­      - e.g. (December, 15, 2012) will construct the date December 15th, 2012

       If the constructor is unable to recognize the string argument as a valid month name,
       then it will issue a console error message: 

       Invalid month name: the Date was set to 1/1/2000.
       (with a newline at the end).

       If the day argument is invalid for the given month (but the month name was valid),
       then the constructor will handle this error in the same manner as the other
       parameterized constructor. 

       This constructor will recognize both "december" and "December"
       as month name.
   */
   Date(const string &mn, unsigned d, unsigned y);


   /* Outputs to the console (cout) a Date exactly in the format "3/1/2012". 
      Does not output a newline at the end.
   */
   void printNumeric() const;


   /* Outputs to the console (cout) a Date exactly in the format "March 1, 2012".
      The first letter of the month name is upper case, and the month name is
      printed in full - January, not Jan, jan, or january. 
      Does not output a newline at the end.
   */
   void printAlpha() const;

 private:

   /* Returns true if the year passed in is a leap year, otherwise returns false.
   */
   bool isLeap(unsigned y) const;


   /* Returns number of days allowed in a given month
      -  e.g. daysPerMonth(9, 2000) returns 30.
      Calculates February's days for leap and non-­leap years,
      thus, the reason year is also a parameter.
   */
   unsigned daysPerMonth(unsigned m, unsigned y) const;

   /* Returns the name of a given month
      - e.g. name(12) returns the string "December"
   */
   string name(unsigned m) const;

   /* Returns the number of a given named month
      - e.g. number("March") returns 3
   */
   unsigned number(const string &mn) const;
};

Private Member Functions

The functions declared private above, isLeap, daysPerMonth, name, number, are helper functions - member functions that will never be needed by a user of the class, and so do not belong to the public interface (which is why they are "private"). They are, however, needed by the interface functions (public member functions), which use them to test the validity of arguments and construct valid dates. For example, the constructor that passes in the month as a string will call the number function to assign a value to the unsigned member variable month.

isLeap: The rule for whether a year is a leap year is:

(year % 4 == 0) implies leap year

except (year % 100 == 0) implies NOT leap year

except (year % 400 == 0) implies leap year

So, for instance, year 2000 is a leap year, but 1900 is NOT a leap year. Years 2004, 2008, 2012, 2016, etc. are all leap years. Years 2005, 2006, 2007, 2009, 2010, etc. are NOT leap years.

Output Specifications

Read the specifications for the print function carefully. The only cout statements within your Date member functions should be:

1. the "Invalid Date" warnings in the constructors

2. in your two print functions

Required main function to be used:

Date getDate();

int main() {

   Date testDate;
   testDate = getDate();
   cout << endl;
   cout << "Numeric: ";
   testDate.printNumeric();
   cout << endl;
   cout << "Alpha:   ";
   testDate.printAlpha();
   cout << endl;

   return 0;
}

Date getDate() {
   int choice;
   unsigned monthNumber, day, year;
   string monthName;

   cout << "Which Date constructor? (Enter 1, 2, or 3)" << endl
      << "1 - Month Number" << endl
      << "2 - Month Name" << endl
      << "3 - default" << endl;
   cin >> choice;
   cout << endl;

   if (choice == 1) {
      cout << "month number? ";
      cin >> monthNumber;
      cout << endl;
      cout << "day? ";
      cin >> day;
      cout << endl;
      cout << "year? ";
      cin >> year;
      cout << endl;
      return Date(monthNumber, day, year);
   } else if (choice == 2) {
      cout << "month name? ";
      cin >> monthName;
      cout << endl;
      cout << "day? ";
      cin >> day;
      cout << endl;
      cout << "year? ";
      cin >> year;
      cout << endl;
      return Date(monthName, day, year);
   } else {
      return Date();
   }
}