Answer:
a-The present value of revenue in the first year is $61,085.92.
b-The total time it would take to pay for its price is 2.44 years of 29.33 months.
Explanation:
a-
Let the function of the revenue earned is given as
![S(t)=\left \{ {{66000t+38000} {\ \ 0The present value is given as [tex]PV=\int\limits^a_b {S(t)e^{-rt}} \, dt](https://tex.z-dn.net/?f=S%28t%29%3D%5Cleft%20%5C%7B%20%7B%7B66000t%2B38000%7D%20%7B%5C%20%5C%200%3C%2Fp%3E%3Cp%3EThe%20present%20value%20is%20given%20as%20%3C%2Fp%3E%3Cp%3E%5Btex%5DPV%3D%5Cint%5Climits%5Ea_b%20%7BS%28t%29e%5E%7B-rt%7D%7D%20%5C%2C%20dt)
Here
- a and b are the limits of integral which are 0 and 1 respectively
- r is the rate of interest which is 5% or 0.05
- S(t) is the function of value which is
![S(t)=\left \{ {{66000t+38000} {\ \ 0So the equation becomes[tex]PV=\int\limits^0_1 {S(t)e^{-0.05t}} \, dt\\PV=\int\limits^{0.5}_0 {(66000t+38000)e^{-0.05t}} \, dt+\int\limits^{1}_{0.5}{(71000)e^{-0.05t}} \, dt\\PV=\int\limits^{0.5}_0 {(66000t)e^{-0.05t}} \, dt+\int\limits^{0.5}_0 {(38000)e^{-0.05t}} \, dt+\int\limits^{1}_{0.5}{(71000)e^{-0.05t}} \, dt\\PV=8113.7805+18764.4669+34207.6751\\PV=61085.9225](https://tex.z-dn.net/?f=S%28t%29%3D%5Cleft%20%5C%7B%20%7B%7B66000t%2B38000%7D%20%7B%5C%20%5C%200%3C%2Fli%3E%3C%2Ful%3E%3Cp%3ESo%20the%20equation%20becomes%3C%2Fp%3E%3Cp%3E%5Btex%5DPV%3D%5Cint%5Climits%5E0_1%20%7BS%28t%29e%5E%7B-0.05t%7D%7D%20%5C%2C%20dt%5C%5CPV%3D%5Cint%5Climits%5E%7B0.5%7D_0%20%7B%2866000t%2B38000%29e%5E%7B-0.05t%7D%7D%20%5C%2C%20dt%2B%5Cint%5Climits%5E%7B1%7D_%7B0.5%7D%7B%2871000%29e%5E%7B-0.05t%7D%7D%20%5C%2C%20dt%5C%5CPV%3D%5Cint%5Climits%5E%7B0.5%7D_0%20%7B%2866000t%29e%5E%7B-0.05t%7D%7D%20%5C%2C%20dt%2B%5Cint%5Climits%5E%7B0.5%7D_0%20%7B%2838000%29e%5E%7B-0.05t%7D%7D%20%5C%2C%20dt%2B%5Cint%5Climits%5E%7B1%7D_%7B0.5%7D%7B%2871000%29e%5E%7B-0.05t%7D%7D%20%5C%2C%20dt%5C%5CPV%3D8113.7805%2B18764.4669%2B34207.6751%5C%5CPV%3D61085.9225)
So the present value of revenue in the first year is $61,085.92.
b-
The time in which the machine pays for itself is given as
The present value is set equal to the value of machine which is given as
$160,000 so the equation becomes:
So the total time it would take to pay for its price is 2.44 years of 29.33 months.
Answer:
Closing value of inventory = $357 for 21 units
Explanation:
As for the provided information we have,
Under FIFO method we know,
FIFO means First In First Out, under this the goods bought at earliest are sold earliest.
That means first opening inventory is sold, then the inventory purchased at the earliest.
Now we have,
Opening Inventory = 27 units @ $17 = $459
Purchases:
Aug 5 22 units @ $16 = $352
Aug 12 26 units @ $17 = $442
Provided 54 units are sold on Aug 15, that means, opening inventory of 27 units, 22 units bought on Aug 5, and 54 - 27 - 22 = 5 units from purchases on Aug 12.
Therefore, after sale units left = 26 - 5 = 21 units
Thus, closing value of inventory = $357 for 21 units
The principal duhhh dumb add jhit
Answer:
C Protection against inflation
Explanation:
As we know that there are three functions of money i.e.
1. Unit of account
2. Store of value
3. Medium of exchange
There is only 3 functions of money that are shown above
So the protection against inflation would not be considered for the same
And, these 3 would represent the functions of money and can be treated as the unit of account, store of value and the medium of exchange
Hence, the option c is correct
