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3 years ago

When are you required to operate a motorboat or pwc at 30 miles per hour or less?

1 answer:

3 0

When are you required to operate a motorboat or PWC at 30 miles per hour or less the operator should operate between one-half hour after sunset and one hour before sunrise. It is illegal to operate a motorboat or PWC in excess of 30 miles per hour without any permission at any time from a half hour after sunset until one hour before sunrise when on waters on the state and therefore be accused by exceeding night speed limit. In addition, when the operator sees buoys or sign, they specify a boating restricted area established to protect the safety of the public and property. In these areas, a vessel may not progress at a speed greater than essential to preserve steering.

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Van has plans to go to an opera and already has a $100 nonrefundable, nonexchangeable, and nontransferable ticket. Now Amy, whom

yan [13]

Answer:

The right choice is "3 Correctly ignored a sunk cost"

Explanation:

As the ticket to the opera was already bought and it is nonrefundable, nonexchangeable, and nontransferable; whether Van decides to go to the opera or to go to the party with Amy; he has incurred $100 cost of ticket which can not be recovered in any manner.

The ticket cost in this question is categorized as sunk cost - cost that incurred in the past and will be remained the same regardless of any future actions. Thus, this type of cost should be ignored when making decision for the future.

So, "3 Correctly ignored a sunk cost" is the correct choice.

4 0

3 years ago

A company is considering buying a new piece of machinery. A 10% interest rate will be used in the computations. Two models of th

JulsSmile [24]

Answer:

Machine I

capitalized cost: 230,271.28

EAC: $ 27,047.58

Machine II

EAC: $ 27,377.930

As Machine I cost per year is lower it is better to purchase that one.

Annual deposits to purchase Machine I in 20 years: $ 1,396.770

return of machine I with savings of 28,000 per year: 10.51%

Explanation:

WE calculate the present worth of each machine and then calculate the equivalent annual cost:

MACHINE 1

Operating cost:

$C \times \frac{1-(1+r)^{-time} }{rate} = PV\\$

C 18,000

time 20

rate 0.1

$18000 \times \frac{1-(1+0.1)^{-20} }{0.1} = PV\\$

PV $153,244.1470

Salvage value:

$\frac{Maturity}{(1 + rate)^{time} } = PV$

Maturity $20,000.0000

time 20.00

rate 0.1

$\frac{20000}{(1 + 0.1)^{20} } = PV$

PV 2,972.87

Total: -80,000 cost - 153,244.15 annual cost + 2,972.87 salvage value:

Total: 230,271.28

$PV \div \frac{1-(1+r)^{-time} }{rate} = C\\$

Present worth $(230,271.28)

time 20

rate 0.1

$-230271.28 \div \frac{1-(1+0.1)^{-20} }{0.1} = C\\$

C -$ 27,047.578

Fund to purchase in 20 years:

$FV \div \frac{(1+r)^{time} -1}{rate} = C\\$

FV $80,000.00

time 20

rate 0.1

$80000 \div \frac{(1+0.1)^{20} -1}{0.1} = C\\$

C $ 1,396.770

IF produce a 28,000 savings:

we must solve using a financial calcualtor for the rate at which the capitalized cost equals 28,000

$PV \div \frac{1-(1+r)^{-time} }{rate} = C\\$

PV $230,271.28

time 20

rate 0.105126197

$230271.28 \div \frac{1-(1+0.105126197287798)^{-20} }{0.105126197287798} = C\\$

C $ 28,000.000

rate of 0.105126197 = 10.51%

<u>Machine II</u>

100,000 cost

25,000 useful life

15,000 operating cost during 10 years

20,000 for the next 15 years

Present value of the operating cost:

$C \times \frac{1-(1+r)^{-time} }{rate} = PV\\$

C 15,000

time 10

rate 0.1

$15000 \times \frac{1-(1+0.1)^{-10} }{0.1} = PV\\$

PV $92,168.5066

$C \times \frac{1-(1+r)^{-time} }{rate} = PV\\$

C 20,000

time 15

rate 0.1

$20000 \times \frac{1-(1+0.1)^{-15} }{0.1} = PV\\$

PV $152,121.5901

in the timeline this is at the end of the 10th year we must discount as lump sum for the other ten years:

$\frac{Maturity}{(1 + rate)^{time} } = PV$

Maturity $152,121.5901

time 10.00

rate 0.1

$\frac{152121.590126167}{(1 + 0.1)^{10} } = PV$

PV 58,649.46

salvage value

$\frac{Maturity}{(1 + rate)^{time} } = PV$

Maturity $25,000.0000

time 25.00

rate 0.1

$\frac{25000}{(1 + 0.1)^{25} } = PV$

PV 2,307.40

Total cost: 100,000 + 92,168.51 + 58,649.46 - 2,307.40 = $248,510.57

$PV \div \frac{1-(1+r)^{-time} }{rate} = C\\$

PV $248,510.57

time 25

rate 0.1

$248510.57 \div \frac{1-(1+0.1)^{-25} }{0.1} = C\\$

C $ 27,377.930

4 0

4 years ago

The textile industry in indonesia has traditionally been one of the largest employers in the manufacturing sector. increased pro

kiruha [24]

Other major Indonesian industries reported greater increases in production value than those reported by the Indonesian textile industry.

<h3><u>Explanation:</u></h3>

In a particular period of time, the monetary value that are obtained by a country due to the production and selling of goods and services is referred to as GDP. It is an important factor that determines the growth rate of a country in an economic framework.

In the example given, it can inferred that the textile industry of Indonesia is considered as the major contributor of GDP because the textile industries has largest amount of employees. But at the end GDP decreased and this is because other major industries in Indonesia has major contribution and production values than the textile industries.

6 0

3 years ago

Suppose that the government decides to issue tradable permits for a certain form of pollution.

ddd [48]

Answer:

1) False

2) True

Explanation:

6 0

3 years ago

A company incurred the following costs: Selling and administrative expenses: $45,000; Direct materials: $15,000; Income tax expe

Westkost [7]

Answer:

$50,000

Explanation:

<em>Manufacturing cost is sum of direct material plus direct labour and manufacturing overhead</em>

Direct material is the cost of all materials directly consumed for production purpose.

Direct labour is the cost of labour hours used for directly for production purpose

Manufacturing cost = 15,000+30,000 + 5000

=$50,000

5 0

3 years ago