Pleasee help 15 points

Laker Company reported the following January purchases and sales data for its only product. Date Activities Units Acquired at Co

Serjik [45]

Complete Question,

Q) Laker Company reported the following January purchases and sales data for its only product. Date Activities Units Acquired at Cost Units sold at Retail Jan. 1 Beginning inventory 140 units @ $ 6.00 = $ 840 Jan. 10 Sales 100 units @ $ 15 Jan. 20 Purchase 60 units @ $ 5.00 = 300 Jan. 25 Sales 80 units @ $ 15 Jan. 30 Purchase 180 units @ $ 4.50 = 810 Totals 380 units $ 1,950 180 units Required: The Company uses a periodic inventory system. For specific identification, ending inventory consists of 200 units, where 180 are from the January 30 purchase, 5 are from the January 20 purchase, and 15 are from beginning inventory. Determine the cost assigned to ending inventory and to cost of goods sold using

(a) specific identification

(b) weighted average

(c) FIFO

(d) LIFO

Answer:

Specific Identification Method:-

Ending Inventory = $925.

Costs of goods Sold = $1025.

Weighted Average Method:-

Ending Inventory = $1026.

Costs of goods Sold = $924

FIFO Method:-

Ending Inventory = $910.

Costs of goods Sold = $1040

LIFO Method:-

Ending Inventory = $1140.

Costs of goods Sold = $=810

<h3>Step-by-step explanation:</h3>

(a) Specific identification method:

Ending inventory:

180 units * $4.50 = $810

5 units * $5 = $25

15 units * $6 = $90

200 units * 0 = 0

$810 + $25 + $90 + 0= $925

Cost of goods sold:

125 units * $6 = $750

55 units * $5 = $275

180 units * 0 = 0

$750 + $275 + 0 = $1025

(b) Weighted average method:

Weighted average rate/unit = $1950 / 380 units = $5.131579

Cost of goods sold:

180 units * $5.131579 = $924

Ending inventory:

200 units * $5.131579 = $1026.

(c) FIFO

Cost of goods sold:

140 units * $6 = $840

40 units * $5 = $200

180 units * 0 = 0

$840 + $200 + 0 = $1040

Ending inventory:

20 units * $5 = $100

180 units * $4.50 = $810

200 units * 0 = 0

$100 + $810 + 0 = $910

(d) LIFO

Cost of goods sold:

180 units * $4.5 = $810

180 units * $0 = 0

$810

Ending inventory:

140 units * $6 = $840

60 units * $5 = $300

200 units * $0 = $0

$840 + $300 = $1140

4 0

3 years ago

Let f(x)=e^x+3 show that f^-1(x)=ln x-3 can someone explain this to me

andrey2020 [161]

Explanation:

For a function ...

y = f(x)

the inverse function is ...

x = f(y)

To write that in the form y=f^-1(x), you need to solve for y.

__

For your function f(x) = e^x +3 this looks like ...

x = e^y +3

x -3 = e^y . . . . . subtract 3

ln(x -3) = y . . . . . take the natural log

So, we can write ...

f^-1(x) = ln(x -3)

_____

Attached is a graph of these functions. As you might guess from the interchange of x and y, a function and its inverse are mirror images of each other across the line y=x.