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tigry1 [53]
1 year ago
11

If we let the domain be all animals, and S(x) = "x is a spider", I(x) = " x is an insect", D(x) = "x is a dragonfly", L(x) = "x

has six legs", E(x, y ) = "x eats y", then the premises be
"All insects have six legs," (∀x (I(x)→ L(x)))
"Dragonflies are insects," (∀x (D(x)→I(x)))
"Spiders do not have six legs," (∀x (S(x)→¬L(x)))
"Spiders eat dragonflies." (∀x, y (S(x) ∧ D(y)) → E(x, y)))
The conditional statement "∀x, If x is an insect, then x has six legs" is derived from the statement "All insects have six legs" using _____.

a. existential generalization
b. existential instantiation
c. universal instantiation
d. universal generalization
Mathematics
1 answer:
Sonbull [250]1 year ago
4 0

Answer:

The conditional statement "∀x, If x is an insect, then x has six legs" is derived from the statement "All insects have six legs" using "a. existential" generalization

Step-by-step explanation:

In predicate logic, existential generalization is a valid rule of inference that allows one to move from a specific statement, or one instance, to a quantified generalized statement, or existential proposition. In first-order logic, it is often used as a rule for the existential quantifier in formal proofs.

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MrMuchimi

Answer:

500 calories

Step-by-step explanation:

divide the 1250 calories by the initial five servings and you get 250 then multiply that by the 2 servings they are asking for and you get 500 calories

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2 years ago
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La fuerza necesaria para evitar que un auto derrape en una curva varía inversamente al radio de la curva y conjuntamente con el
Vika [28.1K]

Answer:

768 libras de fuerza

Step-by-step explanation:

Tenemos que encontrar la ecuación que los relacione.

F = Fuerza necesaria para evitar que el automóvil patine

r = radio de la curva

w = peso del coche

s = velocidad de los coches

En la pregunta se nos dice:

La fuerza requerida para evitar que un automóvil patine alrededor de una curva varía inversamente con el radio de la curva.

F ∝ 1 / r

Y luego con el peso del auto

F ∝ w

Y el cuadrado de la velocidad del coche

F ∝ s²

Combinando las tres variaciones juntas,

F ∝ 1 / r ∝ w ∝ s²

k = constante de proporcionalidad, por tanto:

F = k × w × s² / r

F = kws² / r

Paso 1

Encuentra k

En la pregunta, se nos dice:

Suponga que 400 libras de fuerza evitan que un automóvil de 1600 libras patine alrededor de una curva con un radio de 800 si viaja a 50 mph.

F = 400 libras

w = 1600 libras

r = 800

s = 50 mph

Tenga en cuenta que desde el

F = kws² / r

400 = k × 1600 × 50² / 800

400 = k × 5000

k = 400/5000

k = 2/25

Paso 2

¿Cuánta fuerza evitaría que el mismo automóvil patinara en una curva con un radio de 600 si viaja a 60 mph?

F = ?? libras

w = ya que es el mismo carro = 1600 libras

r = 600

s = 60 mph

F = kws² / r

k = 2/25

F = 2/25 × 1600 × 60² / 600

F = 768 libras

Por lo tanto, la cantidad de fuerza que evitaría que el mismo automóvil patine en una curva con un radio de 600 si viaja a 60 mph es de 768 libras.

7 0
2 years ago
Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro
bearhunter [10]

The question is incomplete. Here is the complete question:

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.95 probability that he will hit it. One day, Samir decides to attempt to hit  10 such targets in a row.

Assuming that Samir is equally likely to hit each of the 10 targets, what is the probability that he will miss at least one of them?

Answer:

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Step-by-step explanation:

Let 'A' be the event of not missing a target in 10 attempts.

Therefore, the complement of event 'A' is \overline A=\textrm{Missing a target at least once}

Now, Samir is equally likely to hit each of the 10 targets. Therefore, probability of hitting each target each time is same and equal to 0.95.

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We know that the sum of probability of an event and its complement is 1.

So, P(A)+P(\overline A)=1\\\\P(\overline A)=1-P(A)\\\\P(\overline A)=1-0.5987\\\\P(\overline A)=0.4013=40.13\%

Therefore, the probability of missing a target at least once in 10 attempts is 40.13%.

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Answer:

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