The First-In, First-Out inventory costing method assumes that the inventory items ordered first are the first sold. This is ideal for goods that are highly perishable, for example fresh milk. Since no figures or dates are given, we will assume that the month is March 2019 and use any figures to make the example.

Date Item Quantity of stock Cost Price

01 Opening stock bought on Feb 28 10 100

05 Sale of 5 goods (cost is $10 each) (5) 50

15 Purchase of stock (20 goods at $20 each) 20 400

25 Sale of 15 goods (15) 250

(5 at $10 each & 10 at $20 each)

31 Closing Stock 10 200

(20 goods bought on 15th - 10 goods sold on 25th)

The quantity on hand at the end of the month is 10 units.

Total cost of goods on hand at end of the month = 10 units * $20 = 200.

Total cost of goods purchased during the month = $20 * 20 units = $400

Total cost of goods sold during the month = [($10 *5) + ($10 * 5)+ ($20 * 10)] = $200

6 0

4 years ago

Through: (2,2), parallel to y=x+42)

shtirl [24]

Answer:

y=x

Step-by-step explanation:

We want a line parallel to y =x+42

The slope of y =x+42 is 1

Parallel lines have the same slope

We have the slope m=1 and a point (2,2)

We can use point slope form to write a line

y-y1 = m(x-x1)

y-2 = 1(x-2) point slope form

Changing to slope intercept form

y-2 = x-2

Add 2 to each side

y-2+2 = x-2+2

y=x

3 0

4 years ago

A bird was perched on a tree at a height of 15 feet. The bird then saw food on a branch below and dropped down 61/3 feet. What i

natka813 [3]

The current height of the bird is -16/3

The bird perched at a height of 15 feet

The bird dropped to a height of 61/3 feet when it saw a food

The current height can be calculated as follows

= 15 - 61/3

= 45-61/3

= -16/3

Hence the current height of the bird is -16/3

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5 0

2 years ago

In a previous year, 51% of females aged 15 and older lived alone. A sociologist tests whether this percentage is different tod

Yanka [14]

Answer:

Given:

n = 550

Po = 51% = 0.51

P0 - 1 = 0.51 - 1 = 0.49

p' = $\frac{287}{550}$ = 0.5218

np0(1 - p0) = 550*0.51(1-0.51)

= 137.4 ≥ 100

This means the sample size is greater than 5% of the population size. and the sample can be reasonably assumed to be random, the requirement for testing hypothesis are satisfied.

1) The null and alternative hypotheses:

H0 : p0 = 0.51

H1 : p0 ≠ 0.51

The test statistic Z,

$Z = \frac{p' - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

$Z = \frac{0.5218 - 0.51}{\sqrt{\frac{0.51(1-0.51)}{550}}}$

Z = 0.55

This is two tailed test.

The pvalue for Z =0.55

Pvalue = 0.5824

3) Decision:

Since pvalue, 0.5824 is greater than significance level, we fail to reject null hypothesis H0.

Conclusion:

Do not reject Upper H0. There is not sufficient evidence at the alpha equals 0.01 level of significance to conclude that the proportion of females who are living alone has changed.

8 0

3 years ago