Simplify it plzzzzz and thx
1 answer:
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Answer:
0.095163
Step-by-step explanation:
given that a starter motor used in a space vehicle has a high rate of reliability and was reputed to start on any given occasion with probability .99999
Here we find that for any start, there are exactly two outcomes either success or failure.
Also each start is independent of the other since p = 0.99999 for succss is given constant.
Thus X no of successes is binomial with p = 0.99999 and n =10000
If Y is taken as failure then Y is binomial with p' = 0.00001 and n =10000
Required probability
= the probability of at least one failure in the next 10,000 starts
= 1-P(no failure in 10000 starts)
=
Answer:
The maximum number of works that he can write while staying in his time budget is 24.
21 poems and 3 short stories
Step-by-step explanation:
In order to solve this problem we must first determine what our variables are. In this case it's the number of poems and short stories he can write.
p = # of poems
s = # of short stories
Next, we must build our objective function which will represent the total number of works he can write.
N=p+s
where N is the number of works.
Next, we must write the constrains based on the information provided by the problem.
The problem tells us that it takes him 30 hours to write a poem and 70 hours to write a short story and that he has 840 hours available to write them, so that constrain will be the following:
it also tells us that he wants to write at least 4 poems and 3 short stories so there we have our other two constrains.
once we got our constrains we can go ahead and graph them to see how they will behave. (See attached picture)
In the graph p is the horizontal axis and s is the vertical axis.
On the graph we can see a polygon that is formed by the restriction. The vertices of the polygon will represent the optimal conditions for this linear programming problem. There are three optimal solutions there, so we need to test them to see which will return the greatest number of works he can write while keeping the given conditions.
Option 1:
4 poems and 3 short stories
N=4+3
N= 7 works
Option 2:
4 poems and 10 short stories
N=4+10
N=14 works
Option 3:
21 poems and 3 short stories
N=21+3
N=24 works
So the optimal solution will be given by option 3 with 21 poems and 3 short stories.
