The accounts receivable subsidiary ledger is a book of accounts that provides supporting detail for Accounts Receivable.
Answer:
Book value= $51,875
Explanation:
Giving the following information:
Purchase price= $80,000
Salvage value= $5,000
Useful life= 8 years
<u>First, we need to calculate the annual depreciation under the straight-line method:</u>
Annual depreciation= (original cost - salvage value)/estimated life (years)
Annual depreciation= (80,000 - 5,000) / 8
Annual depreciation= $9,375
<u>Now, we can determine the book value at the end of 2019:</u>
Book value= purchase price - accumulated depreciation
Book value= 80,000 - (9,375*3)
Book value= $51,875
Answer and Explanation:
The answer is attached below
Answer:
group purchasing organization (GPO)
Explanation:
A Group Policy Object (GPO) is set of standards that offer Active Directory (AD) manager granular permissions over objective configuration, involving accounts, operating systems, programs, as well as other AD items. GPOs are also used to control the Active Directory system centrally and to customize it. The interpretation of code parameters will also included in this.
A group purchasing organization (GPO) in United States is an organization created to exploit a group of companies ' buying power to receive discounts from suppliers based on the mutual purchase power of GPO members '
<span>25 years: No Payment, but total is 250000
6 months earlier. Payment of "P". It's value 1/2 year later is P(1+0.03)
6 months earlier. Payment of "P". It's value 1 year later is P(1+0.03)^2
6 months earlier. Payment of "P". It's value 1½ years later is P(1+0.03)^3
6 months earlier. Payment of "P". It's value 2 years later is P(1+0.03)^4
</span><span>We need to recognize these patterns. Similarly, we can identify the accumulated value of all 50 payments of "P". Starting from the last payment normally is most clear.
</span>
<span>P(1.03) + P(1.03)^2 + P(1.03)^3 + ... + P(1.03)^50
That needs to make sense. After that, it's an algebra problem.
P[(1.03) + (1.03)^2 + (1.03)^3 + ... + (1.03)^50]
</span>
P(<span><span>1.03−<span>1.03^51)/(</span></span><span>1−1.03) </span></span>= <span>250000</span>
