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Methodical content.

NEW QUESTION: 1
Heniser Pet Foods는 X.와 Y의 두 가지 제품을 생산합니다.
제품 X와 Y는 각각 미화 30 달러와 50 달러입니다. 각 제품은 재료 A와
B. 제품은 6 파운드의 재료 A와 12 파운드의 재료 B를 사용합니다. 제품 Y는 12 파운드의 재료 A와 8 파운드의 재료 B를 사용합니다. 회사는 1,200 파운드의 재료 A와 1,760 파운드의 재료 B 만 구입할 수 있습니다. 제조 할 제품의 최적 혼합은 다음과 같습니다.
A. 40 단위의 X 및 120 단위의 Y.
B. 0 단위 X 및 100 단위 Y.
C. 146 단위 및 Y 단위 0 단위.
D. 120 단위 및 40 단위 Y.
Answer: D
Explanation:
Linear programming is a technique used to maximize a contribution margin function or to minimize a cost function, subject to constraints such as scarce resources or minimum/maximum levels of production. Thus, linear programming is often used for planning resource allocations. In this problem, the equation to be maximized, called the objective function, is: U3 $30)K+ $50Y. This equation is to be maximized subject to the constraints on materials. The two constraint functions are:
Material A: 6X + 12Y < 1,200 Material B: I 2X + 8Y < 1,760
One way to solve this problem is to graph the constraint lines and determine the feasible area. The optimal production level is at an extreme point within the feasible area. The graph shows that a production level of 120 units of and 40 units of Y is a feasible production level that maximizes the contribution margin.

NEW QUESTION: 2

A Frequency
1 500,000
2 500,000
6 7,000,000
8 3,000

A. data new;
set temp;
if a in (1, 2) then
b = 'Medium';
else if a = 8 then
b = 'Small';
else if a = 6 then
b = 'Large';
run;
B. data new;
set temp;
if a = 8 then
b = 'Small ';
else if a in(1, 2) then
b = 'Medium';
else if a = 6 then
b = 'Large';
run;
C. data new;
set temp;
if a = 6 then
b = 'Large ';
else if a in (1, 2) then
b = 'Medium';
else if a = 8 then
b = 'Small';
D. data new;
set temp;
if a = 6 then
b = 'Large ';
if a in (1, 2) then
b = 'Small';
run;
Answer: C

NEW QUESTION: 3
Which of the following shows relationship between market yield and time to maturity?
A. Yield curves
B. Smart curves
C. Bezier curves
D. Filling curves
Answer: A