Table/indexed.
Let's look at the three options and see what their advantages and disadvantages are:
Contiguous - In this scheme, the file is stored in contiguous blocks of the disk. It allows for easy random access of the data, but requires a contiguous sequence of blocks large enough to handle the entire file. Since the size of the file specified in this question varies quite a bit over it's lifespan, you're either going to be wasting a lot of space by having an allocation large enough to handle the maximum sized file, or the file will need to be copied whenever it grows and "bumps" into a file that was allocated after it. Because of this, this method is not the best.
Linked - The file is stored as a single, or double linked list of file blocks. This allows for the file to grow or shrink as needed, using only the amount of space needed for the file. Unfortunately, this storage scheme doesn't allow for random access of the file contents and the file can only be accessed sequentially. The question for this problem doesn't specify how the file is being accessed, so as long as random access isn't required, then this would be a reasonable allocation scheme. But I'm assuming that random access will be required, in which case, this scheme isn't ideal.
table/indexed - In this scheme, some disk blocks are used as tables to point to other disk blocks that actually contain the file data. It's almost as fast as contiguous allocation for random access of the file contents, yet allows for the growth and shrinkage of a file like linked allocation. As such, it handles all use cases at a relatively minor cost in total storage required. So this would be the most appropriate allocation scheme since the file access behavior wasn't specified in this question.
A photocopier cost $105,000 when new and has accumulated depreciation of $96,000. if the business discards this plant asset, the result is a loss of 9,000.
During the asset's anticipated useful life, depreciation is allocated in order to charge a fair percentage of the depreciable amount in each accounting period. Amortization of assets with predetermined useful lives is included in depreciation. Depreciation enables businesses to recoup the cost of an item at the time of acquisition. Instead of collecting the full cost of an asset right away, the technique enables businesses to do so during the asset's lifecycle. This enables businesses to replace current assets with the necessary quantity of revenue in the future.
Subtract the asset's cost from its salvage value (what you anticipate it to be worth at the end of its useful life) to determine depreciation using the straight-line technique. The outcome is the amount or depreciable basis.
Depreciation = asset's cost - salvage value
Depreciation = $105,000 - $96,000
Depreciation = $9,000
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Cassidy's approximate monthly payment stands at $1420. if Cassidy lives planning to obtain a loan from her bank for $210,000 for a new home.
<h3>What is the payment monthly?</h3>
The monthly payment is the quantity paid per month to pay off the loan in the time period of the loan. When a loan is taken out it isn't only the top amount, or the original payment loaned out, that needs to be repaid, but also the good that accumulates.
<h3>What is a loan amortization schedule?</h3>
It is described as the systematic method of representing loan payments according to the time in which the principal amount and interest exist mentioned in a list manner
It is given that:
- Cassidy lives planning to obtain a loan from her bank for $210,000 for a new home.
- A fixed annual interest rate of 2.7% compounded monthly for 15 years.
The formula is:
Plug all the values in the above formula:
$1420.
Hence,
Cassidy's approximate monthly payment stands at $1420.
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Answer: $20,478.78
Explanation:
In 14 years the investment will be,
Gold
10,000/2 = 5000
Then use the compound interest formula
5000 * (1+0.07)^ 14 = $12,892.67
For Certificates of Deposits.
Use the Compound interest formula
Rate and period are in years. Convert to semi annual basis.
3%/ 2 = 1.5%
14 * 2 = 28 periods
= 5000 ( 1+ 0.015) ^ 28
= $7,586.11
Add both
=$12,892.67 + $7,586.11
= $20,478.78
Answer:
200
Explanation:
Base on the scenario been described in the question, the position required if the portfolio has a beta 1 is been calculated as follows .
number of contracts required is
Number of contract =10,000,000/(500×100)
Number of contract =10,000,000/50,000
Number of contract =200.
A long put position is needed because the contracts must provide a positive payoff when the market reduces.
