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cluponka [151]
3 years ago
13

How much would nick have over the span of 20 years if he put $1000 in his savings account with 10% of simple interest

Mathematics
1 answer:
Ilya [14]3 years ago
7 0

Step-by-step explanation:

use the simple interest formula

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Help pleaseeeeeeeeeeeee
eduard

Step-by-step explanation:

17.0 there you go I hope this helped

4 0
3 years ago
Read 2 more answers
Evaluate 81/4-615/32
Kaylis [27]

Answer:33/32

Step-by-step explanation:

81/4 - 615/32

32•4=128 to get a common denominator

Then whatever you do to the top you do to the bottom.

81•32=2,592

Then you would do

615•4=2,460

Now you have

2,592/128 - 2,460/128

Now you subtract

2,592-2460=132

You keep the denominator

132/128 and then simplify by dividing both numbers by four and you get

33/32

4 0
3 years ago
What is an equation of a line, in point-slope form, that passes through (1, −7) and has a slope of −2/3 ? (:
lord [1]
A.y+7=2/3(x-1) that is the point slope.
of -2/3

6 0
4 years ago
Read 2 more answers
There are currently 4,000 Rhinos left in the world and they are going extinct at a rate of 3.7% each year. How many Rhinos will
cupoosta [38]

Answer:

2744 Rhinos will be left after 10 years

Step-by-step explanation:

This question has to do with exponential decay rate. We solve this using the formula:

P(t) = Po(1 - r)^t

Where P(t) = Amount left after t years

Po = Initial amount

r = Decay rate = 3.7% = 0.037

t = time in years = 10 years

P(t) = 4000(1 - 0.037)¹⁰

P(t) = 4000 × 0.963¹⁰

P(t) = 4000 × 0.6859032667003232

P(t) = 2743.6130668

Approximately = 2744 Rhinos will be left after 10 years

3 0
3 years ago
On October 1, 2014, Hess Company places a new asset into service. The cost of the asset is $80,000 with an estimated 5-year life
Ede4ka [16]

Answer:

$43,200

Step-by-step explanation:

Data provided in the question:

Cost of the asset = $80,000

Useful life of the machine = 5 years

Salvage value at the end of useful life = $20,000

Now,

Using the double declining method of depreciation

Annual depreciation rate = 2 ×  [1 ÷ useful life ]

= 2 × [ 1 ÷ 5 ]

= 2 × 0.2

= 0.4 or 40%

thus,

The depreciation from October 1, 2014 to December 31, 2014

= Annual Depreciation rate × duration × Book value for 2014

= 0.4 × 3 months × $80,000

= 0.4 × 0.25 year × $80,000

= $8,000

Book value for 2015

= Cost - depreciation till December 31, 2014

= $80,000 - $8,000

= $72,000

Therefore,

Depreciation for the year 2015

= Annual Depreciation rate  × Book value for 2015

= 0.4 × $72,000

= $28,800

Therefore,

the  book value of the plant asset on the December 31, 2015

= Book value for 2015 - Depreciation for the year 2015

= $72,000 - $28,800

= $43,200

4 0
3 years ago
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