The __________ rate tells you how much one unit of currency is worth when converted to another currency.

If two tables do not already have relationships, how can they be related in the Design view?

Irina18 [472]

Answer:C

Explanation:e2020

5 0

3 years ago

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Unexpectedly high inflation tends to hurt ________ most

a_sh-v [17]

Unexpectedly high inflation tends to hurt lenders the most. When lenders lend money, it is valuable , but the amount of money that must be returned to him/her is fixed. Over time, the value of the money keeps depreciating and finally when the borrower does return the money, the value decreases to a very small amount, which is not worth much. For example, let's say a borrower borrows money from a lender to buy a car. With time, the value of money depreciated so much that when the borrower finally returns the money, the same amount of money is not even worth buying a box a matches!

5 0

4 years ago

A process of admission that does not review an applicants academic qualifications is called ? A. early admission

Marrrta [24]

I think the answer is B

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4 years ago

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Kinda running back on my payments anyone got spare change -apple pay numbers below god bless you all

Mandarinka [93]

Ummmmmmmmmmmmmmmmmmmmmm

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3 years ago

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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