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3 years ago

Consider a bond that costs $1,000 and pays an $80 interest payment each year.

Business

1 answer:

Anna007 [38]3 years ago

7 0

Answer:

The interest rate for this bond is 8% per annum.

Explanation:

Given that,

a bond that costs $1,000 and pays an $80 interest each year.

To find the rate of interest, we use the following formula is

I=Prt

Here P = principal= $1,000

I=interest= $80

t=time= 1 year

∴80 = 1000×r×1

$\Rightarrow r=\frac{80}{1000}$

⇒r = 0.080

⇒r= 8%

The yield for this bond is 8% per annum.

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Jenner works for a mountain bike manufacturing company. His company is being sued by hundreds of customers who have been injured

liberstina [14]

The law suit that The customers are going to give here is based on the product liability.

<h3>What is a product liability?</h3>

This is a suit that is made against a company due to the fact that they allowed a defective good to be bought by a consumer.

The company is being sued due to the fact that the customers are injured fron the defective bicycle.

Read more on product liability here: brainly.com/question/25754997

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2 years ago

It costs $34,000 to retrofit the gasoline pumps at a certain filling station so the pumps can dispense E85 fuel (85% ethanol a

victus00 [196]

Answer:

The investor will pay up the rereofitted pumps in a period of 22.52 months.

Explanation:

<em><u>First,</u></em> we solve for the amount of profit generate per month:

21,000 gallons a month x $0.09 per gallon = $1,890

Now, we calcualte the time at which an monthly income of 1890 discounted at 2% per month matches a present value of 34,000

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

C $1,890.00

time n

rate 0.02

PV $34,000.0000

$1890 \times \frac{1-(1+0.02)^{-n} }{0.02} = 34000\\$

$(1+0.02)^{-n}= 1-\frac{34000\times0.02}{1890}$

$(1+0.02)^{-n}= 0.64021164$

We use logarithmics properties to solve for n:

$-n= \frac{log0.64021164021164}{log(1+0.02)$

-22.52006579

n = 22.5200 = 22 and a half month.

7 0

3 years ago

Constant cost industries:

adoni [48]

Answer:

The correct answer to the following question will be Option C.

Explanation:

Constant cost industries seem to be a sector wherein the proportion of units produced as well as manufacturing costs every unit maintains the very same irrespective including its amount of manufacturing or rise in population. Which doesn't use input data in the appropriate amount to influence the rates of that same components by a shift in industry revenue.
This doesn't even use inputs in such amounts that perhaps the costs of that same inputs will be influenced by a change in business production.

The other choices are not linked to an industry of this kind. Therefore the clarification above is correct.

7 0

3 years ago

A new building that costs $1,400,000 has a useful life of 10 years and a scrap value of $100,000. Using straight-line depreciati

xz_007 [3.2K]

Answer:

V = $1,400,000 - $130,000t

Explanation:

Data provided in the question:

Cost of the new building = $1,400,000

Useful life = 10 years

Scrap value = $100,000

Now,

using the straight line method

Annual depreciation = [ Cost - Scrap value ] ÷ Useful life

= [$1,400,000 - $100,000 ] ÷ 10

= $130,000

Value of building = Cost of the building - Depreciation for 10 years

V = $1,400,000 - [ Annual depreciation × Time ]

V = $1,400,000 - $130,000t

4 0

3 years ago

Your birthday is next week and instead of other presents, your parents promised to give you $2,200 in cash. Since you have a par

vfiekz [6]

Answer:

Interest revenue from the CD 470.04

Explanation:

we will calcualte the future value of the CD and from there calculate the interest:

$Principal \: (1+ r)^{time} = Amount$

Principal 2,200.00

time 8.00 (2 years x 4 quarter per year)

rate 0.02450 (9.8% divided by 4 quarter per year)

This divisions and multiplication are done to make time and rate be express i nthe same metric.

$2200 \: (1+ 0.0245)^{8} = Amount$

Amount 2,670.04

Now, we calculate interest revenue:

Amount - Principal

2,670.04 - 2,200 = 470.04

3 0

3 years ago