Answer:
0.11
Step-by-step explanation:
The events are independent, so:
P(A and B) = P(A) × P(B)
P = 0.5 × 0.22
P = 0.11
Answer:
64
Step-by-step explanation:
If the mean is 15, the sum of 5 numbers is:
Minimum value for the first four numbers would be:
Then the fifth number is:
So the maximum difference is:
The recursive formula for the given sequence as required in the task content is; f(n) = f (n - 1) - 50.
It follows from the task content that the recursive formula for the given sequence is to be determined.
By observation, the sequence is an arithmetic progression and the common difference, d can be evaluated as;
d = 750 - 800 = 800 - 850 = 850 - 900 = -50
Also, since the recursive formula for an arithmetic sequence takes the form;
f(n) = f (n - 1) + d.
Hence, since the recursive formula as required is;
f(n) = f (n - 1) - 50.
Read more on recursive formula;
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Answer:
f(1)=900
f(n)=f(n-1)-50if n>1
Step-by-step explanation:
this is the correct
Use rules of inference to prove that the following conclusion follows from these hypotheses:
C : ∃x (p(x) ∧ r(x))
Clearly label the inference rules used at every step of your proof.
2. Consider the following hypotheses:
H1 : ∀x (¬C(x) → ¬A(x)) H2 : ∀x (A(x) → ∀y B(y)) H3 : ∃x A(x)
Use rules of inference to prove that the following conclusion follows from these hypotheses:
C : ∃x (B(x) ∧ C(x))
Clearly label the inference rules used at every step of your proof.
3. Consider the following predicate quantified formula:
∃x ∀y (P (x, y) ↔ ¬P (y, y))
Prove the unsatisfiability of this formula using rules of inference.
Answer:
See deductions below
Step-by-step explanation:
1)
a) p(y)∧q(y) for some y (Existencial instantiation to H1)
b) q(y) for some y (Simplification of a))
c) q(y) → r(y) for all y (Universal instatiation to H2)
d) r(y) for some y (Modus Ponens using b and c)
e) p(y) for some y (Simplification of a)
f) p(y)∧r(y) for some y (Conjunction of d) and e))
g) ∃x (p(x) ∧ r(x)) (Existencial generalization of f)
2)
a) ¬C(x) → ¬A(x) for all x (Universal instatiation of H1)
b) A(x) for some x (Existencial instatiation of H3)
c) ¬(¬C(x)) for some x (Modus Tollens using a and b)
d) C(x) for some x (Double negation of c)
e) A(x) → ∀y B(y) for all x (Universal instantiation of H2)
f) ∀y B(y) (Modus ponens using b and e)
g) B(y) for all y (Universal instantiation of f)
h) B(x)∧C(x) for some x (Conjunction of g and d, selecting y=x on g)
i) ∃x (B(x) ∧ C(x)) (Existencial generalization of h)
3) We will prove that this formula leads to a contradiction.
a) ∀y (P (x, y) ↔ ¬P (y, y)) for some x (Existencial instatiation of hypothesis)
b) P (x, y) ↔ ¬P (y, y) for some x, and for all y (Universal instantiation of a)
c) P (x, x) ↔ ¬P (x, x) (Take y=x in b)
But c) is a contradiction (for example, using truth tables). Hence the formula is not satisfiable.
Please answer with explanation and process
Answer:
Hey!
69/8 as a mixed number is...
8 5/8!
To get this answer, simply divide 69 by 8, then subtract the WHOLE NUMBER from 69 and then the left over number is the numerator over 8
Hope this helps!
Answer:
8 5/8
Step-by-step explanation:
Answer:
x = -7
Step-by-step explanation: