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GREYUIT [131]
3 years ago
9

Part C: Test the four possible vertices that you found in the objective function in part A. Use those values to determine which

set of values maximizes the objective function.
I included part A as a reference in the picture. I only need the answer to Part C, which is listed above. Thank you for the help!

Mathematics
1 answer:
RideAnS [48]3 years ago
8 0

9514 1404 393

Answer:

  • C(5, 0, 55) = 430
  • C(5, 30, 25) = 400
  • C(15, 30, 15) = 420
  • C(45, 0, 15) = 510 . . . maximum

Step-by-step explanation:

Only two of the vertices are listed in this problem statement. The other two are intersections of x+y+z = 60 with x=5 and the constraints on y. We assume that ...

  0 ≤ y ≤ 30

so the missing vertices are ...

  (x, y, z) = (5, 0, 55)  and  (5, 30, 25)

The two given vertices are ...

  (x, y, z) = (15, 30, 15)  and  (45, 0, 15)

Then the objective function values are ...

  C(5, 0, 55) = 9·5 +6·0 +7·55 = 45 +0 +385 = 430

  C(5, 30, 25) = 9·5 +6·30 +7·25 = 45 +180 +175 = 400

  C(15, 30, 15) = 9·15 +6·30 +7·15 = 135 +180 +105 = 420

  C(45, 0, 15) = 9·45 +6·0 +7·15 = 405 +0 +105 = 510

The objective function is maximized at (x, y, z) = (45, 0, 15).

__

Shown in the attachment are the equality constraints. The inequality constraints overlap in the octant closest to the upper front corner in the figure. That is, the feasible region is the section of the orange plane that is above the red plane, left of the black plane, and in front of the blue plane. The semi-transparent purple plane is the maximized objective function. It intersects the orange plane at the lower left vertex, (x, y, z) = (45, 0, 15).

_____

<em>Additional comment</em>

The maximum calorie burn found here includes no aerobics. If an aerobic workout is required to exercise certain muscle groups, a minimum constraint needs to be put on y. This problem statement has no such constraint.

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Step-by-step explanation:

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6 0
3 years ago
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vichka [17]
The number in the three different forms is as follows.

2/5 (fraction), 0.2 (decimal), and 20% (percent).

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6 0
4 years ago
An example problem in a Statistics textbook asked to find the probability of dying when making a skydiving jump.
MArishka [77]

Answer:

(a) 0.999664

(b) 15052

Step-by-step explanation:

From the given data of recent years,  there were about 3,000,000 skydiving jumps and 21 of them resulted in deaths.

So, the probability of death is \frac{21}{3000000}==0.000007.

Assuming, this probability holds true for each skydiving and does not change in the present time.

So, as every skydiving is an independent event having a fixed probability of dying and there are only two possibilities, the diver will either die or survive, so, all skydiving can be regarded as is Bernoulli's trial.

Denoting the probability of dying in a single jump by q.

q=7\times 10^{-6}=0.000007.

So, the probability of survive, p=1-q

\Rightarrow p=1-7\times 10^{-6}=0.999993.

(a) The total number of jump he made, n=48

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=\binom(n,r)p^rq^{n-r}

=\binom(48,48)(0.999993)^{48}(0.000007)^{48-48}

=(0.999993)^{48}=0.999664 (approx)

So, the probability of survive in 48 skydiving is 0.999664,

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Let, total n skydiving jumps required to meet the surviving probability of 0.9.

So, By using Bernoulli's equation,

0.9=\binom {n }{r} p^rq^{n-r}

Here, r=n.

\Rightarrow 0.9=\binom{n}{n}p^nq^{n-n}

\Rightarrow 0.9=p^n

\Rightarrow 0.9=(0.999993)^n

\Rightarrow \ln(0.9)=n\ln(0.999993) [ taking \log_e both sides]

\Rightarrow n=\frac {\ln(0.9)}{\ln(0.999993)}

\Rightarrow n=15051.45

The number of diving cant be a fractional value, so bound it to the upper integral value.

Hence, the total number of skydiving required to meet the 90% probability of surviving is 15052.

3 0
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hodyreva [135]

Answer:

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= (-3/8) x (-3/1)

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8 0
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