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svet-max [94.6K]
3 years ago
9

12) As of August, 2011, Russia has had 21 rocket launch failures out of 743 total launches. To the nearest whole number, how man

y failures can be expected from the next 800 launches?
23
800
743
44
16,800
Mathematics
1 answer:
bogdanovich [222]3 years ago
5 0

Answer:

23

Step-by-step explanation:

It is given that, in Russia there are 21 rocket launch failures out of 743 total launches. We need to find no of failures for the next 800 launches.

21 rocket = 743 launches

1 launch = \dfrac{21}{743}\ \text{failures}

For 800 launches,

=\dfrac{21}{743}\ \text{failures}\times 800\\\\=22.611\ \text{failures}

or

= 23 failures

So, there are 23 failures in next 800 launches.

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Step-by-step explanation:

a) The LP formulation for this problem is:

Objective function (minimize cost):

C=0.50A+0.20B

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C=0.50*0+0.20*0.625=0.125

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C=0.50*0.583+0.20*0.333=0.358

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C=0.50*0.75+0.20*0=0.375

The optimum solution changes. The cost is now 0 pounds of ingredient A and 0.625 pounds of ingredient B. The cost is $0.125 per ration.

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