Answer:
The answer is D. $1,830
Explanation:
FIFO means First in First out.
It is one of the inventory methods along with LIFO(Last in First out), average weighted cost and specific identification.
FIFO literally means the inventory bought first will be the first to be sold. Leaving the last inventories bought as the ending inventory.
In this question, Cost of Sales according to FIFO is:
250 units x $6 = $1,500
30 units at $11 = $330
Total =. $1,830
Therefore, the cost of sales under this method is $1,830
Product Life cycle Management has several stages. The stages determined the growth and low downs of product. During the introduction stage, sales are growing slow(low) and profit is minimal.
- Market Introduction stage is often called the introduction stage that has a low growth rate of sales as the product is said to be recently brought into limelight and consumers may not know much about it.
Most times, firms do experience losses rather than profits during this stage. and so when product is new on the market and small or no profit is made due to high costs and low sales.
Conclusively, the stage gives a lot of opportunities such as low competition in the market.
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Answer:
3 years
Explanation:
Calculation to determine The payback period
Using this formula
Payback period=Capital investment/ Increase cash flows
Let plug in the formula
Payback period=$45,000/$15,000
Payback period=3 years
Therefore The payback period is 3 years
Answer:
Option (d) , Bank 4 offers the highest amount after a year
Explanation:
The total amount from each of the interest rates can be expressed as;
A=P(1+r/n)^nt
where;
A=Future value of investment
P=Initial value of investment
r=Annual interest rate
n=Number of times the interest is compounded annually
t=number of years of the investment
a). Bank 1
P=x
r=6.1%=6.1/100=0.061
n=1
t=assume number of years=1
replacing;
A=x(1+0.061/1)^(1×1)
A=x(1.061)
A=1.061 x
b). Bank 2
P=x
r=6%=6/100=0.06
n=12
t=1
Replacing;
A=x(1+0.06/12)^(12×1)
A=x(1.005)^12
A=1.0617 x
c). Bank 3
P=x
r=6%=6/100=0.06
n=1
t=1
Replacing;
A=x(1+0.06/1)^(1)
A=1.0600 x
d). Bank 4
P=x
r=6%=6/100=0.06
n=4
t=1
A=x(1+0.06/4)^(4×1)
A=x(1+0.015)^4
A=x(1.061)
A=1.0614 x
e). Bank 5
P=x
r=6%=6/100=0.06
n=365
t=1
A=x(1+0.06/365)^(365×1)
A=1.0618
Option (d) , Bank 4 offers the highest amount after a year