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kipiarov [429]
2 years ago
15

21. Explique claramente como el fenotipo es el que favorece o desfavorece los resultados productivos obtenidos tanto en las expl

otaciones agrícolas como en las explotaciones pecuarias.
Engineering
1 answer:
Montano1993 [528]2 years ago
6 0
Los sistemas de producción pecuaria, son considerados como la estrategia social, económica y cultural más apropiada para mantener el bienestar de las comunidades, debido a que es la única actividad que puede simultáneamente proveer seguridad en el sustento diario, conservar ecosistemas, promover la conservación de la
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Technician A says the small base circuit of a transistor controls current flow. Technician B says the small emitter circuit cont
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Answer:

A

Explanation:

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2 years ago
1 UNREAD MESSAGE
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Answer:

triangular trade

Explanation:

triangular trade

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2 years ago
Consider an 8-car caravan, where the propagation speed is 100 km/hour, each car takes 1 minute to pass a toll both. The caravan
melamori03 [73]

Answer:

A. 36 minutes

B. 120 minutes

C.

i. 144 minutes

ii. 984 minutes

D. Car 1 is 1.67km ahead of Cat 2 when Car 2 passed the toll B.

E. 98.33km

Explanation

A.

Given

dAb = 10km

dBc = 10km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

                               

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 0.2 hours

= 24 minutes + 0.2 * 60 minutes

= 24 minutes + 12 minutes

= 36 minutes

B.

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

                               

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 0.2 hours

= 3 minutes + 0.2 * 60 minutes

= 3 minutes + 12 minutes

= 15 minutes

Total End delay for 8 cars = 8 * 15 = 120 minutes

C.

Given

dAb = 100km

dBc = 100km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

i. Cars travelling together

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

                               

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 2 hours

= 24 minutes + 2 * 60 minutes

= 24 minutes + 120 minutes

= 144 minutes

ii. Cars travelling separately

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

                               

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 2 hours

= 3 minutes + 2 * 60 minutes

= 3 minutes + 120 minutes

= 123 minutes

Total End delay for 8 cars = 8 * 123 = 984 minutes

D.

Distance = 100km

Time = 1 min/car

Car 1 is 1 minute ahead of car 2 --- at toll A and B

If car 1 leaves toll B after 10 minutes then cat 2 leaves after 11 minutes

Time delay = 11 - 10 = 1 minute

Distance = time * speed

= 1 minute * 100km/hr

= 1 hr/60 * 100 km/hr

= 100/60

= 1.67km

E.

Given

Distance = 100km

Distance behind = 1.67

Maximum value of dBc = 100km - 1.67km = 98.33km

The maximum distance that can be reached is 98.33km

7 0
3 years ago
Suppose that all the dislocations in 3000 mm3 of crystal were somehow removed and linked end to end. Given 1 m =0.0006214 mile,
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Answer:

please let me know the answer to this because I need to know as well. thanks

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Construct an NFAA foe<br>regular expression<br>labi.abb.cob)​
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Explanation:

Construct an NFA:

Step 1 - Formulate an NFA with Void transitions from its specified regular expression.

Step 2 - Drop and transform the Null Transformation from of the NFA into an analogous DFA.

NDFA corresponding to

RE − 1 (0 + 1)* 0.

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