According to Michael Kremer, large populations c. are a prerequisite for technological advances and higher living standards.
<h3>Who is Michael Kremer?</h3>
Michael Kremer is an American development economist and a Nobel Prize winner for developing an innovative economic theory for poverty alleviation, especially in large populations.
Michael Kremer did not think that large populations disadvantaged the nation, but it could be a factor in increasing the living standards through technological advances.
Thus, according to Michael Kremer, large populations c. are a prerequisite for technological advances and higher living standards.
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Answer:
Amortized loan
Explanation:
An amortized loan is a loan with scheduled periodic payments that are applied to both principal and interest. An amortized loan payment first pays off the relevant interest expense for the period, after which the remainder of the payment reduces the principal.
Interest is calculated based on the most recent ending balance of the loan and the interest amount owed decreases as payments are made. This is because any payment in excess of the interest amount reduces the principal, which in turn, reduces the balance on which the interest is calculated.
Answer: true, there is an equilibrium.
Explanation:
the above game is a pure strategy. A pure strategy occur when the game is strictly determined. the criteria used to determine the pure strategy is maxi-min and min-max. max-min means selecting the least case from all best cases and min-max means selecting the best from all least cases.
Base on the criterion, the least out of all best for max-min is 1 unit and the best out of all least cases for min-max is 1 unit. the max-min is equal to the min-max,hence it strictly determined and therefore a pure strategy.
Answer:
400; 800
Explanation:
Contribution:
Product X:
= Selling price - Variable cost
= 100 - 70
= 30,
Product Y:
= Selling price - Variable cost
= 80 - 40
= 40,
Product Z:
= Selling price - Variable cost
= 25 - 20
= 5
Machine hours required :
Product X:
= Machine time per unit × Monthly demand
= 3 × 300
= 900,
Product Y:
= Machine time per unit × Monthly demand
= 2 × 200
= 400,
Product Z:
= Machine time per unit × Monthly demand
= 1 × 500
= 500
Contribution per machine hour:
Product X = Contribution ÷ Machine time per unit
= 30 ÷ 3
= 10,
Product Y = Contribution ÷ Machine time per unit
= 40 ÷ 2
= 20,
Product Z = Contribution ÷ Machine time per unit
= 5 ÷ 1
= 5
It is highest for Y, so produce maximum amount of Y, then X and then Z
Y needs 400 hrs, we are left with 800 hours, so produce 800 hours of X.
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