-4m = 17+7
-4m = 24
(leave m alone and positive)
-4m/ -4 = 24/-4
m= -6
Hope it helps!
#MissionExam001
Let the number of type A surfboards to be ordered be x and the number of type B surfboards be y, then we have
Minimize: C = 272x + 136y
subject to: 29x + 17y ≥ 1210
x + y ≤ 50
x, y ≥ 1
From the graph of the constraints, we have that the corner points are:
(20, 30), (41.138, 1) and (49, 1)
Applying the corner poits to the objective function, we have
For (20, 30): C = 272(20) + 136(30) = 5440 + 4080 = $9,520
For (41.138, 1): C = 272(41.138) + 136 = 11189.54 + 136 = $11,325.54
For (49, 1): C = 272(49) + 136 = 13328 + 136 = $13,464
Therefore, for minimum cost, 20 type A surfboards and 30 type B surfboards should be ordered.
If the question is 6,372= 35x and you were solving for x the answer would be found by dividing 6372 by 35. this would leave you with 182.1=x
Answer:
$30
Step-by-step explanation:
25% of 40 = 10
40-10 = 30
Since it's not mentioned I assume the Event of tossing a die twice are done simultaneously.
1- What is the theoretical probability that a coin toss results in two heads showing? I guess you mean: The theoretical probability of tossing 2 heads in 2 flips, if so P(1st Head) = 1/2 AND P(2nd Head) = 1/2, then the probability of getting 2 heads simultaneously is P(1st Head AND 2nd Head) = 1/2 x 1/2 = 1/4
2- What is the experimental probability that a coin toss results in two heads showing? In this case, I suggest that you toss a coin several times until 2 heads appear and then you calculate P: Assuming you get 2 heads (out of 10toss), the P(H) =2/10
3- What is the experimental probability that a coin toss results in two tails showing? Same logic as the previous one
4- What is the theoretical probability that a coin toss results in one head and one tail showing? If you get one Head, P(H) =1/2 and if you want a tail in the 2nd toss P(1/2). So P(one Head AND one tail) = 1/2 x 1/2 = 1/4
5- What is the experimental probability that a coin toss results in one head and one tail showing? Same as 3- or 4-
