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lubasha [3.4K]
2 years ago
14

How is a calculation of net worth different from a day-to-day or month-to-month tallying of expenses?

Mathematics
2 answers:
anygoal [31]2 years ago
5 0

When net worth is calculated on daily basis, we add total expenditure of a day on a work sheet or on a sheet of paper or using excel , then we  add the total expense of each day using excel or by calculator or by Using mental Math.

Now, coming to monthly expenses, we add total expenditure of 12 month using excel or on a rough sheet of paper or by using calculator .

For, example, A shopkeeper sells 50 ice cream in a day , making a total expense of $ 240 in a day to earn a profit of $ 250.So we can calculate his day to day expense as well as earning on a excel sheet. Similarly we can find his yearly expense or income by adding his monthly income.


muminat2 years ago
4 0
Net worth includes everything of value you own or could get money for. While month to month expenses is just wages or available spending money
You might be interested in
The following chart shows a store?s records of coat sales over two years. 2 circle graphs. A circle graph titled 2006. Top coats
Mkey [24]

Answer:

<u>The correct answer is A. 16.5%</u>

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly and to calculate the trend:

                   2006      2007  Trend

Top Coats    297        223       -24.92%

Parkas          210        210         + 0%  

Jackets         213        285        +33.80%

Raincoats      137        259        +89.05

Trench coats 103       127         +23.30%

Total               960     1,104        +15%

2. If the trend shown in these graphs stays constant, what percent of the market will parkas occupy in 2008?

Let's calculate the percent of the market occupied by parkas.

In 2006 = 210/960 = 21.88%

In 2007 = 210/1,104 = 19.02%

In 2008 = 210/1,270 = 16.54% (210 + 0 = 210; 1,104 + 15% = 1,270)

<u>The correct answer is A. 16.5%</u>

7 0
2 years ago
Read 2 more answers
6. Problems and Applications Q6 Consider an economy that produces only chocolate bars. In year 1, the quantity produced is 3 bar
natulia [17]

Answer:

Year 1 GDP Deflator is 100%

Year 2 GDP Deflator is 30%

Year 3 GDP Deflator is 14.29%

Inflation Rate between year 2 and year 3 is 50%

The Real GDP growth Rate for Year 2 and year 3 is 110%

Step-by-step explanation:

Year 1  

 Price of chocolate bar is $2 and 3 bars are sold that year so the real GDP is 3 x $2=$6 which we are also given that this year is the nominal base year so the nominal GDP is also $6. GDP is the sum of all market value produced products in an economy. Therefore that’s why we calculated as the price of a chocolate multiplied the number produced. To calculate the GDP Deflator will be as follows:

GDP Deflator= (nominal GDP/Real GDP) x 100

                      = ($6/$6) x 100

                       = 100%  

Year 2

Price of chocolate bars is $4 per bar and 5 bars were produced therefore Real GDP =$4 x 5 = $20, now we will calculate the GDP deflator as we have been told that year  is the nominal year therefore nominal GDP is $6.

GDP Deflator= (nominal GDP/Real GDP) x 100

                       = ($6/$20) x100

                        = 30%

Year 3

Price of chocolate bars is $6 per bar and 7 bars were produced therefore Real GDP =$6 x 7 =$42, now we calculate the GDP deflator as we have been told that year 1 is the nominal year therefore nominal GDP is $6.

GDP Deflator = (nominal GDP/ Real GDP) x 100

                      = ($6/$42)

                       =14.29%

Now we calculate the inflation rate between year 2 and year 3.we use the CPI (consumer price index to get the inflation rate for year 2 ad 3)

Consumer Price Index = (Current price of bar/previous price of bar) x 100 formula for CPI

                      = ($6/$4) x 100= 150%-100%

                         = 50% is the inflation rate as the consumer price gave us a positive value.

Now we compute the real GDP growth rate between year 2 and year 3

Real GDP growth rate = [  (current Real GDP- Previous Real GDP)/Previous Real GDP] x 100

                                      = ($42-$20)/$20

                                       = 110% so real GDP grew by 110% from year 2 to year 3.

8 0
2 years ago
Every day, the number of network blackouts has a distribution (probability mass function)
boyakko [2]
Expected Mean, E(X), is obtained by multiplying each pair of x and its P(x) and add up the answers

E(X) = (0×0.7) + (1×0.2) + (2×0.1) = 0.4

The formula to calculate the variance, Var(X), is given by E(X)² - (E(X))²

E(X²) = (0²×0.7) + (1²×0.2) + (2²×0.1) = 0+0.2+0.4 = 0.6
(E(X))² = (0.4)² = 0.16

Var(X) = 0.6 - 0.16 = 0.44

Translating these answers into the context we have

E(Y) = 0.4×500 = $200
Var(Y) = $110

3 0
2 years ago
The width of a rectangle is the length minus 3 units. The area of the rectangle is 54 units. What is the width, in units, of a r
saul85 [17]

Width of the rectangle is 9 units

Step-by-step explanation:

  • Step 1: Let the width of the rectangle be x. Then the length = x - 3. Find dimensions of the rectangle if its area = 54 sq. units

Area of the rectangle = length × width

54 = x (x - 3)

54 = x² - 3x

x² - 3x - 54 = 0

x² + 6x - 9x - 54 = 0 (Using Product Sum rule to factorize)

x(x + 6) - 9(x + 6) = 0

(x + 6)(x - 9) = 0

x = -6, 9 (negative value is neglected)

x = 9 units

4 0
2 years ago
Solve each of the following initial value problems and plot the solutions for several values of y0. Then describe in a few words
Taya2010 [7]

Answer:

Step-by-step explanation:

Answer:

a) y-8 = (y₀-8)  , b) 2y -5 = (2y₀-5)

Explanation:

To solve these equations the method of direct integration is the easiest.

a) the given equation is

          dy / dt = and -8

         dy / y-8 = dt

We change variables

          y-8 = u

         dy = du

We replace and integrate

           ∫ du / u = ∫ dt

           Ln (y-8) = t

We evaluate at the lower limits t = 0 for y = y₀

          ln (y-8) - ln (y₀-8) = t-0

Let's simplify the equation

           ln (y-8 / y₀-8) = t

           y-8 / y₀-8 =

            y-8 = (y₀-8)

b) the equation is

            dy / dt = 2y -5

            u = 2y -5

            du = 2 dy

            du / 2u = dt

We integrate

             ½ Ln (2y-5) = t

We evaluate at the limits

            ½ [ln (2y-5) - ln (2y₀-5)] = t

            Ln (2y-5 / 2y₀-5) = 2t

            2y -5 = (2y₀-5)

c) the equation is very similar to the previous one

             u = 2y -10

             du = 2 dy

             ∫ du / 2u = dt

             ln (2y-10) = 2t

We evaluate

             ln (2y-10) –ln (2y₀-10) = 2t

               2y-10 = (2y₀-10)

4 0
2 years ago
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