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Lunna [17]
3 years ago
5

NEED THE ANSWER ASAP PLEASE

Mathematics
1 answer:
Irina-Kira [14]3 years ago
7 0

It's a linear program; the extrema will be at the corners.

Let's enumerate them.  4 choose 2 gives 6 possible meets of 4 lines.  One pair are parallel, down to five to try.

4x-y=1 intersects x=0 at y=-1, outside the domain y≥0.

4x-y=1 intersects y=0 at x=1/4, (1/4, 0)

4x-y=1 intersects x=5 at y=19, (5,19)

(0,0) is outside the domain 4(0)-0=0 which isn't ≥1.

(5,0) is a valid corner.

It's a triangular domain.  Three points to try,

C(1/4,0) = 6(1/4) + 2(0) = 3/2

C(5,19) = 6(5) + 2(19) = 68

C(5,0) = 6(5) + 2(0) = 30

Answer: Maximum C=68 at (x,y)=(5,19)

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At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the firs
rewona [7]

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4032 different tickets are possible.

Step-by-step explanation:

Given : At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the first two races. If the first race runs 9 horses and the second runs 8.

To find : How many different tickets are possible ?

Solution :

In the first race there are 9 ways to pick the winner for first and second place.

Number of ways for first place - ^9C_1=9

Number of ways for second place - ^8C_1=8

In the second race there are 8 ways to pick the winner for first and second place.

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6.

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