Answer:
$8,800
Explanation:
Calculation for What is the amount of insurance expense that would appear on the company's income statement for the first year ended December 31
First step is to calculate insurance amount per year
Insurance=$26,400/2 years
Insurance= 13,200
Second step is to calculate the insurance value per months
Insurance value=13,200/12 months
Insurance value=1,100
Now let calculate insurance expense
Insurance expense =$1,100 x 8 months
Insurance expense = $8,800
Note that May 1 to December 31 will give us 8 months
Therefore the amount of insurance expense that would appear on the company's income statement for the first year ended December 31 will be $8,800
Answer:
$60,224
Explanation:
Calculation to determine what The adjusted cash balance should be:
Cash account debit balance+Check printing fee, not yet recorded
Let Plug in the formula
Adjusted cash balance=$60,209+$15
Adjusted cash balance=$60,224
Therefore The adjusted cash balance should be:$60,224
Answer:
False
Explanation:
GDP or gross domestic product value is a measure of the total value of all products and services produced within the boundaries of a country in a given time. It factors all products, regardless of who manufactures them, whether foreigners or locals, men or women. To avoid double-counting, GDP considers finished products only.
In calculating GDP, economists will deduct the cost of imports. The reason is that imports are produced in foreign countries. The value of GDP indicates whether the economy is expanding or contracting. An increase in GDP shows economic growth in the country. An increase in capital goods, human capital, labor force, technology, contribute to economic growth.
The production would be a my a point inside the curve. The curve shows the possibility of producing with all possible materials so inside the curve is representative of one or more of the resources not being used to its full capacity.
Answer:
(a) E(X) = 3
(b) Var(X) = 12.1067
Explanation:
(a) E[X]
E[X]T = E[X]T=A + E[X]T=B + E[X]T=C
= (2.6 + 3 + 3.4)/3
= 2.6 (1/3) + 3(1/3) + 3.4(1/3)
= 2.6/3 + 1 + 3.4/3
= 3
(b) Var (X) = E[X²]−(E[X])²
Recall that if Y ∼ Pois(λ), then E[Y 2] = λ+λ2. This implies that
E[X²] = [(2.6 + 2.6²) + (3 + 3²) + (3.4 + 3.4²)]/3
= (9.36 + 12 + 14.96)/3
= 36.32/3
= 12.1067
Var(X) = E[X²]−(E[X])²
= 12 - 3²
= 12.1067 - 9
= 3.1067
