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Ivanshal [37]
2 years ago
7

Mr. Leghorn lives next door to Mr. Fudd. During hunting season, Mr. Fudd likes to shoot rabbits in his backyard, which activity

he values at $900. The noise from the shooting disturbs Mr. Leghorn and prevents him from taking afternoon naps, which he values at $500. If Mr. Leghorn has the legal right to stop Mr. Fudd from hunting, the socially optimal outcome is for:____________
a) Mr. Fudd to pay Mr. Leghorn between $500 and $900 to continue hunting.
b) Mr. Fudd to pay Mr. Leghorn less than $500 to continue hunting.
c) Mr. Fudd to stop hunting.
d) Mr. Leghorn to pay $500 or less to get Mr. Fudd to stop hunting.
Business
1 answer:
Dvinal [7]2 years ago
3 0

Answer:

A. Mr. Fudd to pay Mr. Leghorn between $500 and $900 to continue hunting.

Explanation:

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Reporting Financial Statement Effects of Bond Transactions Lundholm, Inc., which reports financial statements each December 31,
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Answer:

                            Lundholm, Inc

                            Journal Entries

Date          Account Titles                   Debit          Credit

May 1, 18   Cash                                $500,000

                       Bonds payable                               $500,000

                 (To record the bond issuance)                  

31 Oct, 18  Interest Expenses           $22,500

                 (500000*9%*6/12)

                         Cash                                               $22,500

                 (To record payment of the first semiannual period’s interest)

Nov 1, 19  Bonds payable                  $300,000

                Loss on Bonds                  $3,000

                          Cash                                                $303,000

                 (To record retirement the bonds at 101 on November 1, 2019)

8 0
2 years ago
10. The Wetski Water Ski Company is the world’s largest producer of water skis. As you might suspect, water skis exhibit a highl
xenn [34]

Answer: $ 30,290,000

Explanation:

For this question, the transportation table has been addressed below.

“X” indicates that there will be no possible production in the specific cell.

(i) The total cost of optimum production plan = (50,000 x 50) + (50,000 x 78) + (50,000 x 50) + (50,000 x 75) + (20,000 x 91) + (40,000 x 88) + (50,000 x 50) + (50,000 x 75) + (40,000 x 85) + (50,000 x 50) + (2,000 x 75)

= $ 30,290,000.

Therefore, the cost of the plan will be $30,290,000.

5 0
2 years ago
Birk Co. uses a job order cost system. The following debits (credits) appeared in Birk's work-in-process account for the month o
svp [43]

Answer:

The amount of direct materials charged to Job No. 5 is $5,200.

Explanation:

Work in process, April 30 = Balance + Direct material + Direct labor + Factory overhead - Cost of finished goods

                                            = $4,000 + 24,000 + 16,000 + 12,800 - 48,000

                                             = $8,800

Job No 5 = Work in process, April 30 = $8,800

Job No 5 = Direct material + Direct labor + Factory overhead

$8,800 = Direct material + $2,000 + $1,600 ($2,000 * 80%)

Direct material = $8,800 - $2,000 - $1,600

                         = $5,200

Therefore, The amount of direct materials charged to Job No. 5 is $5,200.

5 0
2 years ago
A repetitive manufacturing firm is planning on level material use. The following information has been collected. Currently, the
Sloan [31]

Answer:

setup cost = $1.75

setup time = 2.625 min

Explanation:

given data

firm operates = 250 days per year

Annual demand  = 22,000

Daily demand  =  88

Daily production  = 250

Desired lot size =  63  (2 hours of production)  

Holding cost   = $40 per unit per year

to find out

setup cost  and setup time

solution

we find first setup cost that is express as

setup cost = \frac{Q^2*H*(1-\frac{d}{p})}{2D}   ......................1

here Q is  Desired lot size and H is  Holding cost and d is  Daily demand and D is Annual demand   and p is  Daily production

put here value

setup cost = \frac{63^2*40*(1-\frac{88}{250})}{2*22000}

setup cost = \frac{2969*40*(0.648)}{44000}

setup cost = $1.75

and

setup time is

setup time = \frac{setup\ cost}{setup\ labor}    ....................2

setup time = \frac{1.75*60min/hr}{40}

setup time = 2.625 min

8 0
2 years ago
(20%) problem 7: your uncle is trying to convince you join his research team at cern by offering you a summer job that pays a wa
Nimfa-mama [501]

Solution:

A)

Expected hourly income in $US  = rate * hourly income in Euro

= 1.2903 * 13 = $16.7739

B)

Time taken from uncle's house to lab = Distance /Speed

Distance is  8.5 km  

Speed is 1mile in 5 minutes  

 Hence converting mile in Kilometer = 1.609344 km in 5 minutes  

 distance traveled in an hour = 1.609344 *12 =19.312128

 Time = 8.5/19.312128  = 0.44013792783 hours = 26.408 minutes

 

C)

Capacity =13 US gallons  

1US gallon = 3.78541 litre

Hence 13 gallons = 13 * 3.78541

=49.21033 litres

 

D)

Gasoline = 1.41 Euro /litre

= 1.2903 * 1.41 = 1.819323 $ /litre

Hence 1 litre = 1.819323 dollar

1 gallon = 3.78541 litre  = 6.88688347743 dollar

Hence price = 6.88688347743 dollar /gallon

E)

we can travel 20 miles in 1 gallon  

= 1.609344*20 km  

capacity of tank =13 gallons  

total distance that can be traveled in 13 gallons = 1.609344*20*13 km = 418.42944 km  

days that can be traveled  = 418.42944/8.5 = 49.2269929412 days = 49 days

3 0
2 years ago
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