Answer:
- 30 days at location I
- 35 days at location II
Step-by-step explanation:
Let x and y represent days of operation of Location I and Location II, respectively. Then we want to minimize the objective function ...
650x +750y
subject to the constraints on drape production:
10x +20y ≥ 1000 . . . . . order for deluxe drapes
20x +50y ≥ 2100 . . . . . order for better drapes
13x +6y ≥ 600 . . . . . . . order for standard drapes
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I find a graphical solution works well for this. The vertices of the feasible solution space are (x, y) = (0, 100), (30, 35), (80, 10), (105, 0). The vertex at which the cost is minimized is
(x, y) = (30, 35)
This schedule will produce exactly the required numbers of deluxe and standard drapes, and 2350 pairs of better drapes, 250 more than required.
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In the attached graph, we have reversed the inequalities so that the solution space (feasible region) is white, not triple-shaded. Minimizing the objective function means choosing the vertex of the feasible region so that the line representing the objective function is as close to the origin as possible.
Answer:
see below
Step-by-step explanation:
We can just add up all of these fractions:
2 1/8 + 7/8 = 2 8/8 = 3
8 5/8 + 4/8 = 8 9/8 = 9 1/8
3 + 9 1/8 = 
Answer:
The answer is "496"
Step-by-step explanation:
It begins at 0000 or finishes at 1111:
Answer:
Step-by-step explanation:
A. It says he will get to sit in the front if the number is even
There are 6 sides, three of which are even numbers (2,4,6). So the probability of rolling an even number is 3/6 = 1/2.
Let's make p = 1/2
Jack will get his way half of the time, so we expect him to get the front seat half of the time
If we're talking about 20 days, then n = 20 and
n*p = 20*(1/2) = 20/2 = 10
meaning that Jack would, on average, get the front seat 10 times out of the 20 total.
B. Similar to A
n=100
The probability is still the same because rolling an odd number (1,3,5) is still going to happen 3/6 = 1/2 of the time.
so: n*p means 100 days*1/2 which is 100/2=50 days
Jill get to sit in the front 50 out of the 100 days.
C. For better accuracy, we can at least increase the number of trials so we have more data in able for better accuracy.
As Jill did 100 hundred, we think that she will get closer to the actual frequency. As larger the number of trials, the closer we'll get to the expected probability.
0<x<1 and 0<y<1
x>0 so x is positive and y>0 so y is also positive.
When you multiply two positive numbers you always get a positive number, so the product of x and y must be positive, or greater than 0.
xy>0 - it must be true
xy<0 - it can't be true
Also when you divide a positive number by a positive number you always get a positive number, so the quotient of x and y must be positive.
x/y<0 - it can't be true
D and E can be true, but don't have to. It depends on the values of x and y. If x>y, then x-y>0 is true and x-y<0 isn't true; if x<y, then x-y>0 isn't true and x-y<0 is true.
Therefore, only A <u>must</u> be true.
