Answer:
$658
Explanation:
Henderson assembled 47 dressers in the week.
Her applicable pay rate is $14 per piece since she assembled over 30 dressers in one week.
her gross pay for the week
=$14 x 47
=$658
<span>The answer is if Jim's marginal revenue is greater than his marginal cost.
Marginal revenue refers to the value that will Jim give to the company if the company decided to employ Jim. (how much profit he will create for the company)
The marginal cost on the other hand refers to the value that company must sacrifice in order to keep him working (the salary and benefit)</span>
Answer:
I. If labor and capital are perfect substitutes in production, the isoquant is a straight, downward-sloping line.
II. If a company needs to use inputs in fixed proportion such that the capital to labor ratio is always 2, the firm's isoquants are L-shaped.
Explanation:
Perfectly substittuable goods have straight downward sloping ICs, and have corner solutions
.
Complementary goods (used in fixed proportions) are L shaped always
, In case of min(x,y) function, the answer is the value of x or y which ever is minimum and not their sum.
Therefore, Only statements I and II are true.
Answer:
a-The present value of revenue in the first year is $61,085.92.
b-The total time it would take to pay for its price is 2.44 years of 29.33 months.
Explanation:
a-
Let the function of the revenue earned is given as
![S(t)=\left \{ {{66000t+38000} {\ \ 0The present value is given as [tex]PV=\int\limits^a_b {S(t)e^{-rt}} \, dt](https://tex.z-dn.net/?f=S%28t%29%3D%5Cleft%20%5C%7B%20%7B%7B66000t%2B38000%7D%20%7B%5C%20%5C%200%3C%2Fp%3E%3Cp%3EThe%20present%20value%20is%20given%20as%20%3C%2Fp%3E%3Cp%3E%5Btex%5DPV%3D%5Cint%5Climits%5Ea_b%20%7BS%28t%29e%5E%7B-rt%7D%7D%20%5C%2C%20dt)
Here
- a and b are the limits of integral which are 0 and 1 respectively
- r is the rate of interest which is 5% or 0.05
- S(t) is the function of value which is
![S(t)=\left \{ {{66000t+38000} {\ \ 0So the equation becomes[tex]PV=\int\limits^0_1 {S(t)e^{-0.05t}} \, dt\\PV=\int\limits^{0.5}_0 {(66000t+38000)e^{-0.05t}} \, dt+\int\limits^{1}_{0.5}{(71000)e^{-0.05t}} \, dt\\PV=\int\limits^{0.5}_0 {(66000t)e^{-0.05t}} \, dt+\int\limits^{0.5}_0 {(38000)e^{-0.05t}} \, dt+\int\limits^{1}_{0.5}{(71000)e^{-0.05t}} \, dt\\PV=8113.7805+18764.4669+34207.6751\\PV=61085.9225](https://tex.z-dn.net/?f=S%28t%29%3D%5Cleft%20%5C%7B%20%7B%7B66000t%2B38000%7D%20%7B%5C%20%5C%200%3C%2Fli%3E%3C%2Ful%3E%3Cp%3ESo%20the%20equation%20becomes%3C%2Fp%3E%3Cp%3E%5Btex%5DPV%3D%5Cint%5Climits%5E0_1%20%7BS%28t%29e%5E%7B-0.05t%7D%7D%20%5C%2C%20dt%5C%5CPV%3D%5Cint%5Climits%5E%7B0.5%7D_0%20%7B%2866000t%2B38000%29e%5E%7B-0.05t%7D%7D%20%5C%2C%20dt%2B%5Cint%5Climits%5E%7B1%7D_%7B0.5%7D%7B%2871000%29e%5E%7B-0.05t%7D%7D%20%5C%2C%20dt%5C%5CPV%3D%5Cint%5Climits%5E%7B0.5%7D_0%20%7B%2866000t%29e%5E%7B-0.05t%7D%7D%20%5C%2C%20dt%2B%5Cint%5Climits%5E%7B0.5%7D_0%20%7B%2838000%29e%5E%7B-0.05t%7D%7D%20%5C%2C%20dt%2B%5Cint%5Climits%5E%7B1%7D_%7B0.5%7D%7B%2871000%29e%5E%7B-0.05t%7D%7D%20%5C%2C%20dt%5C%5CPV%3D8113.7805%2B18764.4669%2B34207.6751%5C%5CPV%3D61085.9225)
So the present value of revenue in the first year is $61,085.92.
b-
The time in which the machine pays for itself is given as
The present value is set equal to the value of machine which is given as
$160,000 so the equation becomes:
So the total time it would take to pay for its price is 2.44 years of 29.33 months.
I would think it could be a 4 ×4
