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ioda
3 years ago
11

Suppose that x ∗ is an optimal basic feasible solution, and consider an optimal basis associated with x ∗ . Let B and N be the s

et of basic and nonbasic indices, respectively. Let I be the set of nonbasic indices i for which the corresponding reduced costs are zero.
Mathematics
1 answer:
ch4aika [34]3 years ago
6 0
If I’m honest I’d love to help but I’m pretty bad at math so
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Suppose you by a 1.25 pound package of ham at $5.20 per pound
marishachu [46]
I suppose I can do that and end up spending 6.50$?

1.25 x 5.20 = 6.50 final amount due.
6 0
3 years ago
Neptune’s average distance from the sun is 4.503 x 10e9 km. Mercury’s average distance from the sun is 5.791 x10e9 km.about how
Igoryamba

Answer:

77.76 times

Step-by-step explanation:

The average distance of Neptune  from the sun

= 4.503  ×  10 ⁹  k m .

and Mercury  =  5.791  ×  10 ⁷ k m .

Hence neptune is (  4.503  ×  10 ⁹) ÷ (5.791 × 10 ⁷  ) times farther from the sun than mercury

i.e.(  \frac{4.503}{5.791} )  × 10⁹⁻⁷ times

=   0.7776  ×  10 ² times

=   77.76  times.

4 0
3 years ago
Since there are 6.6 laps in a mile, how many laps will you have to make to run 5 miles?
mezya [45]

Answer:

33 laps

Step-by-step explanation:

we know that

There are 6.6 laps in a mile

so

using proportion

Find how many laps will you have to make to run 5 miles

so

Let

x ----> the number of laps

\frac{6.6}{1}\frac{laps}{mile}=\frac{x}{5}\frac{laps}{mile} \\ \\x=6.6*5\\ \\ x=33\ laps

4 0
3 years ago
Add parentheses 10+9×3-1×2 equal greater than 75
vitfil [10]
(10 + 9) x (3 - 1) x 2 = +75
19 x 2 x 2
76
76 > 75


5 0
3 years ago
Suppose the speeds of vehicles traveling on a highway are normally distributed and have a known population standard deviation of
erik [133]

Answer:

2.88

Step-by-step explanation:

Data provided in the question

\sigma = Population standard deviation = 7 miles per hour

Random sample = n = 32 vehicles

Sample mean = \bar X = 64 miles per hour

98% confidence level

Now based on the above information, the alpha is

= 1 - confidence level

= 1 - 0.98

= 0.02

For \alpha_1_2 = 0.01

Z \alpha_1_2 = 2.326

Now the margin of error is

= Z \alpha_1_2 \times \frac{\sigma}{\sqrt{n}}

= 2.326 \times \frac{7}{\sqrt{32}}

= 2.88

hence, the margin of error is 2.88

5 0
3 years ago
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