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Simora [160]
3 years ago
12

Luke's choice behavior can be represented by the utility function u(x1,x2)=x1+x2. The prices of x1 and x2 are denoted as p1 and

p2, and his income is m.
Draw at least three indifference curves and find its slope (i.e. MRS). Is the MRS changing depending on the points of (x1,x2) at which it is evaluated, or constant?
Mathematics
1 answer:
Lunna [17]3 years ago
6 0

Answer:

Step-by-step explanation:

hahaha idk the answer fool

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The following represents age distribution of students in an elementary class. Find the mode of the values: 7, 9, 10, 13,
Vitek1552 [10]
The mode of the values listed is 9.
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3 years ago
Please help, I will give thanks and 5 stars. I beg for an answer, since I have worked on this question for 2 hours, yet my effor
lbvjy [14]

Answer:its 150 percent or 3/2

Step-by-step explanation:

You can get an even number 1/3 if you spin it. You can get tails 1/2 the time uf you flip it. You can also get an odd number 2/3 of the time if you spin it. If you add 1/3, 1/2, and 2/3 you get 1.5.

8 0
4 years ago
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Solve the equation: p-5/p = 3/9
ExtremeBDS [4]

Answer;

p= (1 + sqrt. 181) / 6, (1- sqrt. 181) / 6

Hope this helps!

6 0
3 years ago
Which of the following shows 2 + (x + 3y) rewritten using the Associative Property of Addition?
Eduardwww [97]
The answer is B. (2 + x) + 3y.

The associative change happened when they just moved the parenthesis.

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8 0
3 years ago
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An aeroplane X whose average speed is 50°km/hr leaves kano airport at 7.00am and travels for 2 hours on a bearing 050°. It then
Zigmanuir [339]

Answer:

(a)123 km/hr

(b)39 degrees

Step-by-step explanation:

Plane X with an average speed of 50km/hr travels for 2 hours from P (Kano Airport) to point Q in the diagram.

Distance = Speed X Time

Therefore: PQ =50km/hr X 2 hr =100 km

It moves from Point Q at 9.00 am and arrives at the airstrip A by 11.30am.

Distance, QA=50km/hr X 2.5 hr =125 km

Using alternate angles in the diagram:

\angle Q=110^\circ

(a)First, we calculate the distance traveled, PA by plane Y.

Using Cosine rule

q^2=p^2+a^2-2pa\cos Q\\q^2=100^2+125^2-2(100)(125)\cos 110^\circ\\q^2=34175.50\\q=184.87$ km

SInce aeroplane Y leaves kano airport at 10.00am and arrives at 11.30am

Time taken =1.5 hour

Therefore:

Average Speed of Y

=184.87 \div 1.5\\=123.25$ km/hr\\\approx 123$ km/hr (correct to three significant figures)

(b)Flight Direction of Y

Using Law of Sines

\dfrac{p}{\sin P} =\dfrac{q}{\sin Q}\\\dfrac{125}{\sin P} =\dfrac{184.87}{\sin 110}\\123 \times \sin P=125 \times \sin 110\\\sin P=(125 \times \sin 110) \div 184.87\\P=\arcsin [(125 \times \sin 110) \div 184.87]\\P=39^\circ $ (to the nearest degree)

The direction of flight Y to the nearest degree is 39 degrees.

7 0
3 years ago
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