B. To help determine how much inventory to keep in stock of each item in the outlet, I would consider two factors. The first one is the <u>certain amount of each item type </u>in the first week - determining the number of each items input in the stock to be the original data to compare. The second is the <u>number of items sold and still on stock</u>. This would help determine invetory to keep in stock.
B. If I was running a store, I would prefer to use the pick up at store buying method. There would be cashier at the register to check the products and bills as well as receive money from the customers. This is the traditional and still always the most common methods. As the fact that people nowadays still like shopping in physical store, this is definitely the most effective method.
C. The inventory control method I would use when operating a store is to adopt modern technology like bar code to control the inventory stock. By using this technology, I would need to consider the <u>bar code</u> for each type and establishing <u>software</u> to automatically input the information about the number of items sold and in stock. With inputting the number of total items at the beginning and giving each item its corresponding bar code, the sales or number of stock would be tracked by the software.
D. An example of two commodities to be displayed together at store is <u>pencil and rubber</u>. Pencil and rubber undoubtedly are complementary products of each other. Complementary products are made to be used together. Each item requires the other for their complete uses. This is also the case of Rubber and Pencil and each item would not be fully used without the presence of the other.
The difference between the monthly payment of R and S is equal to $48.53 by following the compound interest formula. Thus, Loan R's monthly loan amount is greater than Loan S.
<h3>What is a Compound interest loan?</h3>
Combined interest (or compound interest) is the loan interest or deposit calculated based on both the original interest and accrued interest from earlier periods.
![\rm\,For\,R\\\\P = \$\,17,550\\r\,= 5.32\%\\Time\,= n= 7\,years\\Amount\,paid= [P(1+\dfrac{r}{100\times12})^{n\times12} ]\\=[ 17,550 (1+\dfrac{5.32}{100\times12})^{7\times12} ]\\= [ 17,550 (\dfrac{12.0532}{12})^{84} ]\\\\= [ 17,550 (1.00443^{84} ]\\\\= \$ 25,440.48\\\\Total\,monthly\,payment = \rm\,\dfrac{25,440.48}{84}\\\\= \$\, $302.86\\\\](https://tex.z-dn.net/?f=%5Crm%5C%2CFor%5C%2CR%5C%5C%5C%5CP%20%3D%20%5C%24%5C%2C17%2C550%5C%5Cr%5C%2C%3D%205.32%5C%25%5C%5CTime%5C%2C%3D%20n%3D%207%5C%2Cyears%5C%5CAmount%5C%2Cpaid%3D%20%5BP%281%2B%5Cdfrac%7Br%7D%7B100%5Ctimes12%7D%29%5E%7Bn%5Ctimes12%7D%20%5D%5C%5C%3D%5B%2017%2C550%20%281%2B%5Cdfrac%7B5.32%7D%7B100%5Ctimes12%7D%29%5E%7B7%5Ctimes12%7D%20%5D%5C%5C%3D%20%5B%2017%2C550%20%28%5Cdfrac%7B12.0532%7D%7B12%7D%29%5E%7B84%7D%20%5D%5C%5C%5C%5C%3D%20%20%5B%2017%2C550%20%281.00443%5E%7B84%7D%20%5D%5C%5C%5C%5C%3D%20%5C%24%2025%2C440.48%5C%5C%5C%5CTotal%5C%2Cmonthly%5C%2Cpayment%20%3D%20%5Crm%5C%2C%5Cdfrac%7B25%2C440.48%7D%7B84%7D%5C%5C%5C%5C%3D%20%5C%24%5C%2C%20%24302.86%5C%5C%5C%5C)
![\rm\,For\,S =\\\\P=\,\$ 15,925\\r\,= 6.07\%\\T=n= 9\,years\\\\Amount\,paid\,= [P(1+\dfrac{r}{100\times12})^{n\times12} ]\\\\\= [15,925(1+\dfrac{0.0607}{12})^{9\times12} ]\\\\\\= [15,925(1+\dfrac{0.0607}{12})^{108} ]\\\\=[15,925(1.7247.84)} ]\\\\\= \$27,467.19\\\\Total\,monthly\,payment =\dfrac{\rm\,\$\,27,469.19}{108}\\\\= \$ 254.326\\\\](https://tex.z-dn.net/?f=%5Crm%5C%2CFor%5C%2CS%20%3D%5C%5C%5C%5CP%3D%5C%2C%5C%24%2015%2C925%5C%5Cr%5C%2C%3D%206.07%5C%25%5C%5CT%3Dn%3D%209%5C%2Cyears%5C%5C%5C%5CAmount%5C%2Cpaid%5C%2C%3D%20%5BP%281%2B%5Cdfrac%7Br%7D%7B100%5Ctimes12%7D%29%5E%7Bn%5Ctimes12%7D%20%5D%5C%5C%5C%5C%5C%3D%20%5B15%2C925%281%2B%5Cdfrac%7B0.0607%7D%7B12%7D%29%5E%7B9%5Ctimes12%7D%20%5D%5C%5C%5C%5C%5C%5C%3D%20%5B15%2C925%281%2B%5Cdfrac%7B0.0607%7D%7B12%7D%29%5E%7B108%7D%20%5D%5C%5C%5C%5C%3D%5B15%2C925%281.7247.84%29%7D%20%5D%5C%5C%5C%5C%5C%3D%20%5C%2427%2C467.19%5C%5C%5C%5CTotal%5C%2Cmonthly%5C%2Cpayment%20%3D%5Cdfrac%7B%5Crm%5C%2C%5C%24%5C%2C27%2C469.19%7D%7B108%7D%5C%5C%5C%5C%3D%20%5C%24%20254.326%5C%5C%5C%5C)
The difference between the monthly payment of R and S is equal to $48.53.
Hence, Loan R's monthly payment is greater than the loan's monthly payment by $48.53
To learn more about Compound interest, refer to the link:
brainly.com/question/14331235
Answer:
$2,800
Explanation:
An interest only loan represents a type of loan offer where a borrower is only expected to pay the interest either for some of the term of the loan as agreed or for all of the terms of the loan. However, the principal amount that is collected remains constant all through the agreed interest -only period.
Since the loan obtained by John's Auto Repair is Interest Only, it means that the principal of $35,000 remains constant.
Hence, in the 8th year, John is expected to pay only the interest for the period =
0.08 x $35,000
= $2,800
Answer: option C
Explanation: THIS CAN BE REPRESENTED AS FOLLOWS :-
If we eliminate the product there would be no sales, no variable expenses and therefore, no contribution.
sales = nil
-variable expenses= <u>nil</u>
contribution = nil
- fixed expenses = <u>56,000</u>
NET LOSS = <u> (56000)</u>
.
NOTE :-
Fixed expense = (140,000)*(40%)= 56,000
.
.
Thus increase in loss would be 56000- 50,000=6000
Answer:
Gasoline is a normal good
Explanation:
Normal goods are goods that are goods whose demand increases when income increases and falls when income falls
Inferior goods are goods whose demand falls when income rises and increases when income falls.
Because the demand for gasoline falls when income falls, gasoline is a normal good.
I hope my answer helps you
