Answer:
B) the wages received for the fifth day of work.
Explanation:
Marginal benefit is the increment in benefit generated by an increase by one unit of output. In this situation, the marginal benefit is given by difference in wage of working five days a week from the wage of working four days a week. Therefore, the marginal benefit is the wage received for the fifth day of work.
The answer is alternative B)
<em> 7*|15 80 | =|105 560|</em>
<em> 7*|15 80 | =|105 560||40 100| |280 700|</em>
<em> 7*|15 80 | =|105 560||40 100| |280 700|HERE'S YOUR ANSWER </em>
<em> 7*|15 80 | =|105 560||40 100| |280 700|HERE'S YOUR ANSWER ◌⑅⃝●♡⋆♡MICKZMINNZ♡⋆♡●⑅◌</em>
Answer:
''there will be at most as many POSITIVE rates...''
Explanation:
The measure of investments' rate of return which excludes external factors such as inflation is known as Internal Rate of Return(IRR)
It is used in;
(1). Savings and loans.
(2). Liabilities
(3). Fixed incomes
(4). Private equity and capital management.
(5). Maximizing total present value and so on.
It can be calculate using the formula below:
NPV= C(n)/(1+r)^n = 0
That is internal rate of return can be use in solving NPV = 0.
Therefore, 'With respect to engineering economics and the internal rate of return (IRR), Descartes’ rule of signs indicates there will be at most as many POSITIVE rates of return as there are sign changes in the cash flow profile.''
option d. is the right option
Answer:
a) H0: u = presence of a unit root
HA: u ≠ presence of a unit root ( i.e. stationary series )
b) t stat = -0.064
c) We will reject the Null hypothesis and the next step will be to accept the alternative hypothesis
d) It is not valid to compare the estimated t stat with the corresponding critical value because a random walk is non-stationary while the difference is stationary because it is white noise
Explanation:
<u>a) stating the null and alternative hypothesis</u>
H0: u = presence of a unit root
HA: u ≠ presence of a unit root ( i.e. stationary series )
<u>b) performing the test </u>
critical value = -2.88
T stat = coefficient / std error
= -0.02 / 0.31 = -0.064
c) From the test, the value of T stat > critical value we will reject the Null hypothesis hence the next step will be to accept the alternative hypothesis
d) It is not valid to compare the estimated t stat with the corresponding critical value because a random walk is non-stationary while the difference is stationary because it is white noise
