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damaskus [11]
2 years ago
12

Warranty service, processing of complaints, and costs of litigation are examples of Multiple Choice appraisal costs. internal fa

ilure costs. external failure costs. replacement costs. prevention costs.
Business
1 answer:
zhannawk [14.2K]2 years ago
8 0

Examples of internal failure costs include warranty service and complaint handling. As a result, choice b is accurate.

<h3>What do you mean by internal failure cost?</h3>

Internal failure costs are expenses related to flaws discovered prior to the client receiving the good or service. External failure costs are expenses related to flaws discovered after the client has purchased the good or service.

Internal failure costs are quality expenses related to product flaws found before a product leaves the facility.

Hence, warranty services all are examples of the internal failure cost.

Learn more about internal failure costs:

brainly.com/question/14802565

#SPJ1

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A wireless phone service provider advertises that their average startup fee is $73. Given that their startup fees are $8, $85, $
ipn [44]

Answer:

The reason why it is statisticaly misleading is because, while it is true that the average fee is $73 (the median value in statistical terms), averages are a statistical measure that is very sensitive to extreme values.

That is to say, if a value is very high, or very low, the statistical mean will be biased.

We can see this in the question. Three values are higher than the average, and relatively close: $85, $92, and $107. The third value, however, is way lower, at only $8. This extreme low value alters the median value, making it biased and misleading.

6 0
3 years ago
Currie Company borrowed $13,000 from Sierra Bank by issuing a 10% three-year note. Currie agreed to repay the principal and inte
Luda [366]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Download xlsx
7 0
4 years ago
A bond has a par value of $1,000, a time to maturity of 10 years, and a coupon rate of 8% with interest paid annually. If the cu
bogdanovich [222]

Answer:

4.92%

Explanation:

we have to calculate the market price of the bond in one year from now but in order to do this we have to calculate the yield to maturity:

YTM = {80 + [(1,000 - 750)/10] / [(1,000 + 750)/2] = 105 / 875 = 12%

the market price of the bond in one year is:

PV of face value = $1,000 / 1.12⁹ = $360.61

PV of coupon payments = $80 x 5.3282 (PV annuity factor, 12%, 9 periods) = $426.26

market price one year from now = $786.87

capital gains yield = ($786.87 - $750) / $750 = 4.92%

4 0
3 years ago
Unitech has the following inventory information. July 1 Beginning Inventory 20 units at $19 $ 380 7 Purchases 70 units at $20 1,
rewona [7]

Answer:

B. $600

Explanation:

The average cost method assigns a cost to inventory items based on the total cost of goods purchased (or produced) in a period divided by the total number of items purchased (or produced). Weighted Average Unit Cost is calculated by following formula:

Weighted Average Unit Cost = Total Cost of Inventory /Total Units in Inventory

Total value purchased in July = $1,400+$220 = $1,620

Weighted Average Unit Cost = ($380+$1,620)/100 = $20

Ending inventory = 30 x $20 = $600

Noted: The company did not have date of selling merchandise. In the situation, assuming that the company uses periodic inventory system.

8 0
4 years ago
A truck acquired at a cost of $80,000 has an estimated residual value of $8,000, has an estimated useful life of 200,000 miles,
Alexus [3.1K]

Answer:

a. $72,000

b. $0.36

c. $6,480

Explanation:

a. Depreciation cost = Cost of truck - Residual value

= $80,000 - $8,000

= $72,000

b. The depreciation rate = (Cost of truck - Residual value) ÷ Estimated total production

= ($80,000 - $8,000) ÷ 200,000 miles

= $72,000 ÷ 200,000 miles

= $0.36

c. The units-of-activity depreciation for the year per mile = Driven miles × Depreciation rate

= 18,000 × $0.36

= $6,480

6 0
3 years ago
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