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Dmitry [639]
3 years ago
14

Write the code for invoking a method named sendObject. There is one argument for this method which is of type Customer. Assume t

hat there is a reference to an object of type Customer, in a variable called John_Doe. Use this reference as your argument. Assume that sendObject is defined in the same class that calls it.
Computers and Technology
1 answer:
kodGreya [7K]3 years ago
3 0

Answer:

sendObject(John_Doe);

Explanation:

The above code has been written in Java.

Since the calling class is the same that declares it, to invoke the method, simply call its name with its argument(s) in a pair parentheses. The name of the method is "sendObject" and its argument is a reference to an object of type Customer saved in a variable called "John_Doe". This can be written as follows:

sendObject(John_Doe);

Hope this helps!

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Write a function called mul_time that takes a Time_Elapsed object and a number and returns a new Time_Elapsed object that contai
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# Write a function called mul_time that takes a Time object and a number and

# returns a new Time object that contains the product of the original Time and

# the number.

# Then use mul_time to write a function that takes a Time object that

# represents the finishing time in a race, and a number that represents the

# distance, and returns a Time object that represents the average pace (time

# per mile).

# Current Status: Complete

class Time(object):

   """ represents the time of day.

   attributes: hour, minute, second"""

time = Time()

time.hour = 3

time.minute = 0

time.second = 0

def time_to_int(time):

   minutes = time.hour * 60 + time.minute

   seconds = minutes * 60 + time.second

   return seconds

def int_to_time(seconds):

   new_time = Time()

   minutes, new_time.second = divmod(seconds, 60)

   time.hour, time.minute = divmod(minutes, 60)

   return time

def mul_time(time, multicand):

   time_int = time_to_int(time) * multicand

   new_time = int_to_time(time_int)

   if new_time.hour > 12:

       new_time.hour = new_time.hour % 12

#    print ("New time is: %.2d:%.2d:%.2d"

#    % (new_time.hour, new_time.minute, new_time.second))

   return new_time

# mul_time(time, 2)

def race_stats(time, distance):

   print ("The finish time was %.2d:%.2d:%.2d"

         % (time.hour, time.minute, time.second))

   print "The distance was %d miles" % (distance)

   average = mul_time(time, (1.0 / distance))

   print ("The average is: %.2d:%.2d:%.2d per mile"

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race_stats(time, 3)

7 0
2 years ago
Candy Kane Cosmetics (CKC) produces Leslie Perfume, which requires chemicals and labor. Two production processes are available:
Vilka [71]

Answer:

  1. Divide the resources into three parts using the corresponding process 1, process 1, and process 2 formats to maximize the use of the resources.
  2. Get the expected revenue by calculating the product of the total perfume in ounce and the price of an ounce of perfume.
  3. Increase the advertisement hours of the product.
  4. subtract the advert fee from the generated revenue to get the actual revenue.
  5. subtract the cost of production from the actual revenue to get the actual profit.

Explanation:

The get maximum profit, all the resources must be exhausted in production. The labor is divided into a ratio of 1:1:2 ( which is 5000, 5000, 1000), while the chemical units are in the ratio of 2:2:3 (10000,10000,15000). This would produce in each individual processes; 15000, 15000 and 25000 oz, which is a total of 55000 oz of perfume.

The expected revenue is $275000. If 1000oz from the 55000oz of perfume is sold without advertisement, model Jenny's awareness of the perfume increases the demand by 200oz per hour, therefore, 24hours would field 4800oz demanded, which would only take 270 hours to distribute all remaining perfumes.

The cost of production would be $130000 for labor and chemical resources plus the advert cost of $27000 ( 270 hours by 100) which is a total cost of $157000. The actual profit is $118000 ( $275000 - $157000).

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