PLEASE HELP! What does the graph tell you about spontaneous recovery?

Critically discuss the effectiveness of Cost-Volume-Profit analysis duly supported with empirical evidence.

AnnyKZ [126]

Answer:

Cost-volume-profit (CVP) can be used to determine how changes in costs and volume affect a company's operating income and net income. CVP analysis requires that all the company's costs, including manufacturing, selling, and administrative costs, be identified as either variable or fixed.

A CVP analysis consists of five basic components that include: volume or level of activity, unit selling price, variable cost per unit, total fixed cost, and sales mix.

From the cost-volume-profit analysis, one can determine the sales quantity needed to break even as well as the sales quantity required to earn a desired profit margin .

For instance, given a fixed costs of $4,000 and contribution margin of $20, it becomes possible to determine the volume of sales in order for the entity to break-even (make no profit or loss). The break-even point = $4,000/$20 = 200. This implies that if the entity can sell 200 units with the current level of fixed and variable costs, and selling price, it can make no profit or loss. If more quantity is sold, then the entity can record some profit, and vice versa. Management can use the information provided to decide if production of a product or service can be continued or discontinued if it meets or does not meet the profit goal.

But, the CVP analysis is not always accurate. CVP analysis technique assumes that all costs in the company are completely fixed or completely variable. Fixed costs are costs that do not change with changes in production, such as rent or insurance costs. They are not always completely fixed as they may change periodically, then exhibiting a step fixed costs nature, though in the long run, all costs are variable.

Another issue with CVP analysis is that it is a short run, marginal analysis: it assumes that unit variable costs and unit revenues are constant, which is appropriate for small deviations from current production and sales, and assumes a neat division between fixed costs and variable costs.

Explanation:

Cost-volume-profit (CVP) analysis is a managerial accounting technique to determine how changes in costs and volume affect a company's operating income and net income.

5 0

3 years ago

Explain in detail, how you solved the following problem: The first two terms of a sequence are 10 and 20. If each term after the

KIM [24]

Answer:

$T_{2020} = 15$

Explanation:

Given

$T_1 = 10$

$T_2 = 20$

Each term after the second term is the average of all of the preceding terms

Required:

Explain how to solve the 2020th term

Solve the 2020th term

Solving the 2020th term of a sequence using conventional method may be a little bit difficult but in questions like this, it's not.

The very first thing to do is to solve for the third term;

The value of the third term is the value of every other term after the second term of the sequence; So, what I'll do is that I'll assign the value of the third term to the 2020th term

This is proved as follows;

From the question, we have that "..... each term after the second term is the average of all of the preceding terms", in other words the MEAN

$T_{n} = \frac{\sum T{k}}{n-1} ; where: k = 1 .... n -1$

Assume n = 3

$T_{3} = \frac{T_1 + T_2}{2}$

Multiply both sides by 2

$2 * T_{3} = \frac{T_1 + T_2}{2} * 2$

$2T_{3} = T_1 + T_2$

Assume n = 4

$T_{4} = \frac{T_1 + T_2 + T_3}{3}$

$T_{4} = \frac{(T_1 + T_2) + T_3}{3}$

Substitute $2T_{3} = T_1 + T_2$

$T_{4} = \frac{2T_3 + T_3}{3}$

$T_{4} = \frac{3T_3}{3}$

$T_{4} = T_3$

Assume n = 5

$T_{5} = \frac{T_1 + T_2 + T_3 +T_4}{4}$

$T_{5} = \frac{(T_1 + T_2) + T_3 +(T_4)}{4}$

Substitute $2T_{3} = T_1 + T_2$ and $T_{4} = T_3$

$T_{5} = \frac{2T_3 + T_3 +T_3}{4}$

$T_{5} = \frac{4T_3}{4}$

$T_{5} = \frac{(5-1)T_3}{5-1}$

Replace 5 with n

$T_{n} = \frac{(n-1)T_3}{n-1}$

(n-1) will definitely cancel out (n-1); So, we're left with

$T_{n} = T_3$

Hence,

$T_{2020} = T_3$

Calculating $T_3$

$T_{3} = \frac{10 + 20}{2}$

$T_{3} = \frac{30}{2}$

$T_{3} = 15$

Recall that $T_{2020} = T_3$

$T_{2020} = 15$

3 0

3 years ago

Does low points mean low work or am i outbid by to many questions per point ratio

SpyIntel [72]

You get points by answering other people you can use this currency for homework help, more points = more people that would want to help you out.

5 0

2 years ago

Tulong kailangan tama

vivado [14]

Answer:

1.D

2.P

3.E

4D

5.P

Explanation:

Im not sure..........

8 0

3 years ago

Read 2 more answers

Limit as x approaches 0 of (2e^x-2)/x

Snowcat [4.5K]

Answer:

Explanation:

$\lim_{x \to \ 0} \frac{2e^x-2}{x}$

So first, we have to plug in zero and see if we can evaluate this limit simply from that.

When we plug in zero we get: (2e^0-2)/0

e^0 is 1 so we have 2-2/0 or 0/0. So we have an indeterminate form type 0/0.

This means we have to apply L'Hospital's Rule.

As a reminder L'Hospitals Rule is $\lim_{x\to \ c} \frac{f'(x)}{g'(x)}$

Meaning that we take the derivative of the top and bottom function as the approach some value "c". We can do this with a 0/0 indeterminate form.

So:

The derivative of 2e^x - 2 is just 2e^x

and the derivative of x is 1

So we are left with $\lim_{x \to \ 0} 2e^x$

Plugging in zero we see this gives us 2 as 2(e^0) = 2(1) = 2.

Hence, $\lim_{x \to \ 0} \frac{2e^x-2}{x}$ = 2

3 0

3 years ago