Answer: 
Step-by-step explanation:
Given
There is 4 fluid ounces of red paint which needs to be shared among 5 students
So, division operation is needed to deduce the ounces of red paint each student will get
Each student get
In both cases, we can use a rule of three.
<h2>First part</h2>
Therefore, 4 students correspond to 20% of 20 students in total.
<h2>Second part</h2>
Therefore, 2 doctors correspond to 8% of 25 doctors in total.
Answer:
Described
Step-by-step explanation:
A solution becomes infeasible when no solution exit and which satisfies all the constraints. We will consider two basic types of infeasibility. The 1st we will call continuous infeasibility and the second one is discrete or integer infeasibility. Continuous infeasibility is the one where a non–MIP problem is infeasible. In this case the feasible region defined by the intersecting constraints is empty. Discrete or integer infeasibility is the one where a MIP problem has a feasible relaxation (note that a relaxation of a MIP is the problem we get when we drop the discreteness required on the variables) but the feasible region of the relaxation contains no solution that satisfies the discreteness requirement.
Answer:no they are not independent
Step-by-step explanation:
1) in order for events to be independent the probability of E happening should equal the probability of E happening given that F has already happend
2) P(E)=1/6 (on 6 sided die there is only 1 face that has 3 hence probability of e occurring is 1/6) or .167
3) P(E | F) read as probability of E happening given F has occurred
(Probability of E * probability of F) / total number of outcomes in F
probability of F= 1/6 as there is only one face(3) which will make the sum of the 2 die 6
(1/6)(1/6)/6=0.004
P(E)= .1167
P(E| F) =.004
hence probability aren't equal as making F happen impacts E'E's chances
ANOTHER way to think about this is if E rolls a 3 then the only way to make the sum of the 2 dice equal 3 is if die 2 rolls 3 as well. Hence the pprobability Of the second outcome tht sum of 2 die=6 depends on the number the first die lands
D (2, 5) > (-6, -2)
E (5, 5) > (-3, -2)
F (6, 3) > (-2, -4)
G (5, 1) > (-3, -6)
H (2, 1) > (-6, -6)
I (1, 3) > (-7, -4)
Looking at the transitions all side by side, you could tell they all turn negative so one can assume the transition is a negative.
Therefore, the rule would be:
(x-8, y-7)
