First calculate the amount financed
Amount financed=725−50=675
The formula is
I=(2yc)/(m (n+1))
Solve for c to get
C=(I×m×(n+1))/2y
C=(0.14×675×(24+1))÷(2×12)=98.44
Total of payments=675+98.44=773.44
Monthly payment is
773.44÷24=32.23
Hope it helps!
Answer:
Not impaired because the fair value of the equipment is greater than the carrying value of the asset by $120,000.
Explanation:
Impairment will happen if carrying amount is greater than the fair value of the assets, here the carrying value of the assets is 1,480,000, which is lessor than the fair value of the assets 1,600,000 by 120,000. Hence impairment will not happen so 1st option is correct.
Answer:
1. Repetitive and continuous
2. Call
3. How, what, and when
4. Material problem
Explanation:
The work of Stickley furniture is described as repetitive and continuous. This is because, the company will continue to be producing the same kind of furniture continuous, This will be so because, furniture have a limited number of make. For example, furniture are chairs, tables, cabinet, stool and shelf. These things will be continually produced by the company.
The company will be able to keep track of job status and location through call.
Meanwhile, when the company received an order. The questions they will ask are, what type of furniture they want? When they want it?, How they want it delivered, is it in batches? Where they want it delivered?
Answer:
32
Explanation:
First bounce = 13 / 14 × 10 = 130 /14
using geometric progression where the common ratio = 13/14, the first bound = 130/14
ar^n-1 < 1
substitute the values into the equation
130 /14 × 13/14^(n-1) < 1
(13/14)^n-1 < 1÷ (130/14)
(13/14)^n-1 < 14 / 130
take log of both side
log (13 /14)^n-1 < log ( 14/130)
n-1 log (13 /14) < log ( 14/130)
since log (13/14) negative
n-1 > (log( 14/130)) ÷ ( log (13/14)
n - 1 > 30.07
n > 30.07 + 1 > 31.07
The 32 bounce will the first less than 1 foot
Payroll is your answer.
Payroll is a list that have all employees listed on it as well as the amount they were to be paid during a certain amount of time.
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