Let x represent the number of type A table and y represent the number of type B tables.
Minimize: C = 265x + 100y
Subject to: x + y ≤ 40
25x + 13y ≥ 760
x ≥ 1, y ≥ 1
From, the graph the corner points are (20, 20), (39, 1), (30, 1)
For (20, 20): C = 265(20) + 100(20) = $7,300
For (39, 1): C = 265(39) + 100 = $10,435
For (30, 1): C = 265(30) + 100 = $8,050
Therefore, for minimum cost, 20 of type A and 20 of type B should be ordered.
X+y=425. This is as simple it can be, unless theres more information. x is the distance traveled on the first day and y is the distance traveled on the second day.
For the first roll, you expect that you would get an even number. Since there are six sides of a die, then there are 3 ways to get an even number which are numbers 2, 4 and 6. So the probability of an even number is 3/6 or 1/2.
Next for the second roll, you expect that you should not get a number 2. There are 5 ways occurring which are 1, 3, 4, 5, 6. So the probability for this is 5/6.
Multiply the two probabilities because both probabilities will occur for this event.
P(even, then not 2) = (1/2)(5/6) = 5/12
9514 1404 393
Answer:
birr 80
Step-by-step explanation:
The total number of ratio units is 1+3+5+7 = 16. The largest amount is represented by 7 ratio units, so is 7/16 of the total. The total contributed is ...
16/7 × (birr 35) = birr 80
