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timofeeve [1]
3 years ago
7

How is Demoragan's law used in logic ?

Mathematics
2 answers:
Murrr4er [49]3 years ago
6 0

Answer:

De Morgan's Laws describe how mathematical statements and concepts are related through their opposites. In set theory, De Morgan's Laws relate the intersection and union of sets through complements. In propositional logic, De Morgan's Laws relate conjunctions and disjunctions of propositions through negation.

Mekhanik [1.2K]3 years ago
4 0

Answer:

Step-by-step explanation:

In propositional logic and Boolean algebra, De Morgan's laws are a pair of transformation rules that are both valid rules of inference. ... The rules allow the expression of conjunctions and disjunctions purely in terms of each other via negation.

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Race times at the local monthly 5k run are normally distributed, with a mean time of 32 minutes and a standard deviation of 4 mi
galina1969 [7]

Answer:

68%

Step-by-step explanation:

The mean is 32 minutes, and the standard deviation is 4 minutes.

28 is 1 standard deviation below the mean, and 36 is 1 standard deviation above the mean.

According to the empirical rule, 68% of a normal distribution is between -1 and 1 standard deviations.  95% is between -2 and 2 standard deviations.  99.5% is between -3 and 3 standard deviations.

So the answer is 68%.

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1. Choose the graph that represents the equation y = -x] - 1.
AVprozaik [17]

Answer:

x=−y−1

Step-by-step explanation:

y=−x−1

Step 1: Flip the equation.

−x−1=y

Step 2: Add 1 to both sides.

−x−1+1=y+1

−x=y+1

Step 3: Divide both sides by -1.

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5 0
3 years ago
The curb weight of a new car is 3,347 pounds. How much does the car weigh?
frozen [14]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
Better Products, Inc., manufactures three products on two machines. In a typical week, 40 hours are available on each machine. T
Kaylis [27]

Answer:

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

Step-by-step explanation:

                                Profit $    mach. 1      mach. 2

Product 1     ( x₁ )       30             0.5              1

Product 2    ( x₂ )       50             2                  1

Product 3    ( x₃ )       20             0.75             0.5

Machinne 1 require  2 operators

Machine   2 require  1  operator

Amaximum of  100 hours of labor available

Then Objective Function:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Constraints:

1.-Machine 1 hours available  40

In machine 1    L-H  we will need

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

2.-Machine 2   hours available  40

1*x₁  +  1*x₂   + 0.5*x₃   ≤  40

3.-Labor-hours available   100

Machine 1     2*( 0.5*x₁ +  2*x₂  +  0.75*x₃ )

Machine  2       x₁   +   x₂   +  0.5*x₃  

Total labor-hours   :  

2*x₁  +  5*x₂  +  2*x₃  ≤  100

4.- Production requirement:

x₁  ≤  0.5 *( x₁ +  x₂  +  x₃ )     or   0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

5.-Production requirement:

x₃  ≥  0,2 * ( x₁  +  x₂   +  x₃ )  or    -0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

General constraints:

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

The model is:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Subject to:

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

1*x₁  +  1*x₂   + 0.5*x₃       ≤  40

2*x₁  +  5*x₂  +  2*x₃        ≤  100

0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

-0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

After 6 iterations with the help of the on-line solver AtomZmaths we find

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

6 0
3 years ago
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