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Tanya [424]
3 years ago
12

QV-TV, Inc. provided the following items in its notes to the financial statements for the year-end 2014: Cost of goods sold was

$22 billion under FIFO costing and the inventory value under FIFO costing was $2.1 billion. The LIFO Reserve for year-end 2013 was $0.6 billion and at year-end 2014 it had increased to $0.8 billion.
What is the LIFO inventory value at year-end 2014?

a.$1.9 billion.

b.$2.9 billion.

c.$2.3 billion.

d.$1.3 billion.
Business
1 answer:
Nookie1986 [14]3 years ago
5 0

Answer:

The correct answer is option (d) $1.3 billion.

Explanation:

Given Data:

inventory value under FIFO costing = $2.1 billion

LIFO Reserve for year-end 2013 = $0.6 billion

LIFO Reserve for year-end 2014 = $0.8 billion

LIFO inventory is calculated using the formula;

LIFO inventory value at year-end 2014 = FIFO inventory - LIFO reserve

                                                               =$2.1 billion-$0.8 billion

                                                               = $1.3 billion

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KatRina [158]
Business processes implement value chains or portions of value chains. Each value chain is supported by one or more business process. The information systems are then implemented in order to make the operation <span>run smoothly and productively.</span>
5 0
3 years ago
14. Suppose that the production of $1 million worth of steel in Canada requires $100,000 worth of taconite. Canada’s nominal tar
VMariaS [17]

Answer:

The effective rate of protection for Canada’s steel industry is 21%

Explanation:

The computation of the effective rate is shown below:

Steel percentage = (Production worth of steel) ÷ (Taconite worth)

                             = ($1,000,000) ÷ ($100,000)

                             = 10%

And the tariff rate for steel is 20%

And the taconite percentage is 10%

So, the effective rate would be equal to

= Tariff rate for steel + taconite percentage × steel percentage

= 20% + 10% × 10%

= 20% + 1%

= 21%

7 0
3 years ago
Which of the following government offices help individuals fund their college education?
Doss [256]
D. office of student federal aid
7 0
3 years ago
Suppose that the last four months of sales were 8, 10, 15, and 9 units, respectively. Suppose further that the last four forecas
Wewaii [24]

Answer:

3

Explanation:

Data provided in the question:

Sales for the last four months :

8, 10, 15, and 9 units

Last four forecast of sales:

9, 11, 8 and 12 units

Now,

The mean absolute deviation (MAD) value of these forecast will be calculated as:

MAD = [ ∑|Sales - Forecast sales| ] ÷ [ Total number of forecast ]

or

MAD =  [ |8 - 9| + |10 - 11| + |15 - 8| + |9 - 12| ] ÷ 4

or

MAD = [ 1 + 1 + 7 + 3 ] ÷ 4

or

MAD = 12 ÷ 4

or

MAD = 3

4 0
3 years ago
A university officer wants to know the proportion of registered students that spend more than 20 minutes to get to school. He se
vlada-n [284]

Answer:

1) We need a random sample. For this case we assume that the sample selected was obtained using the simple random sampling method.

2) We need to satisfy the following inequalities:

n\hat p =25*0.52= 13 \geq 10

n(1-\hat p) = 25*(1-0.52) =12 \geq 10

So then we satisfy this condition

3) 10% condition. For this case we assume that the random sample selected n represent less than 10% of the population size N . And for this case we can assume this condition.

So then since all the conditions are satisfied we can conclude that we can apply the normal approximation given by:

p \sim N (\hat p, \sqrt{\frac{\hat p (1-\hat p)}{n}})

So then the answer for this case would be:

a. Yes.

Explanation:

For this case we assume that the question is: If in the experiment described we can use the normal approximation for the proportion of interest.

For this case we have a sample of n =25

And we are interested in the proportion of registered students that spend more than 20 minutes to get to school.

X = 13 represent the number of students in the sample selected that have a time more than 20 min.

And then the estimated proportion of interest would be:

\hat p = \frac{X}{n}= \frac{13}{25}= 0.52

And we want to check if we can use the normal approximation given by:

p \sim N (\hat p, \sqrt{\frac{\hat p (1-\hat p)}{n}})

So in order to do this approximation we need to satisfy some conditions listed below:

1) We need a random sample. For this case we assume that the sample selected was obtained using the simple random sampling method.

2) We need to satisfy the following inequalities:

n\hat p =25*0.52= 13 \geq 10

n(1-\hat p) = 25*(1-0.52) =12 \geq 10

So then we satisfy this condition:

3) 10% condition. For this case we assume that the random sample selected n represent less than 10% of the population size N . And for this case we can assume this condition.

So then since all the conditions are satisfied we can conclude that we can apply the normal approximation given by:

p \sim N (\hat p, \sqrt{\frac{\hat p (1-\hat p)}{n}})

So then the answer for this case would be:

a. Yes.

3 0
3 years ago
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