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frutty [35]
3 years ago
14

A manufacturing operations consists of 12 operations. However, five of the 12 machining operations must be completed before any

of the remaining operations can begin. Within each of these two sets, operations can be completed in any order. How many different production sequences are possible?
Mathematics
1 answer:
shutvik [7]3 years ago
4 0

Answer:

604800 possibilities

Step-by-step explanation:

There are two sets.

The first set is for the first five machining operations. This set has the following operations: A,B,C,D,E

The first operation to be made can be any of them(A,B,C,D,E). The second will be any of the four remaining ones, the third will be any of the three remaining and so on...

So for the first set there are 5*4*3*2*1 = 5! = 120 possibilities.

Now for the second set, the same logic is applied.

There will be 7*6*5*4*3*2*1 = 7! = 5040 possibilities.

Now taking into account all the 12 operations, separated by set, there will be 120*5040 = 604800 possibilities.

You can understand this calculus as that for a simple possibility of the first set, there are 5040 possibilities of the second set. The are 120 possiblities in the first set, so we have the multiplication in the paragraph above.

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Answer:

a) \frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}

b) P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736

Step-by-step explanation:

Part a

We assume that the parachutist lands at random point in the interval (x=A,y=B) we have a continuous random variable X. And the distribution of X would be uniform Y\sim Unif(A,B). And the density function would be given by:

f(x) =\frac{1}{B-A} , A

And 0 for other case.

The operation fails, if parachutist's distance to A is more than five times as much as her distance to B.

So the point P in the interval (A,B) at which the distance to A is exactly 5 times the distance to B is given by:

P= A + \frac{5}{6} (B-A)= \frac{6A +5B -5A}{6}=\frac{A+5B}{6}

And we can find the probability desired like this:

P(d(P,A) \geq 5 d(P,B))= P(\frac{A+5B}{6} < X< B)

And from the cumulative distribution function of X ficen by F(X)\frac{X-A}{B-A} we got:

\frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}

Part b

For this case we assume that X\sim Gamma (2,1)

On this case we assume that \alpha=2, \beta= 1

The density function for the Gamma distribution is given by:

P(X)= \frac{\beta^{\alpha} x^{\alpha-1} e^{-\beta x}}{\gamma(\alpha)}

And on this case we can find the probability using the complement rule like this:

P(X>1) = 1-P(X\leq 1)=0.736

We can solve this problem with the following excel code:

"=1-GAMMA.DIST(1;2;1;TRUE)"

And if we do it by hand we need to do this:

P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736

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