A student is applying to Harvard and Dartmouth. He estimates that he has a probability of .5 of being accepted at Dartmouth and .3 of being accepted at Harvard. He further estimates the probability that he will be accepted by both is .2. What is the probability that he is accepted by Dartmouth if he is accepted by Harvard? Is the event "accepted at Harvard" independent of the event "accepted at Dartmouth"?
Answers
Answer:
Step-by-step explanation:
Let P(H) be the probability that he is accepted by Harvard
Let P(D) be the probability that he is accepted by Dartmouth
Let P(HnD) be the probability that he is accepted by Harvard and Dartmouth
Given data
P(H) = 0.3
P(D) = 0.5
P(DnH) = 0.2
To get the probability that he is accepted by Dartmouth if he is accepted by Harvard, can be gotten using the conditional probability formula.
P(D|H) = P(DnH)/P(H)
P(D|H) = 0.2/0.3
P(D|H) = 2/10/÷3/10
P(D|H) = 2/10×10/3
P(D|H) 2/3
b) The two events are independent if the occurrence of an event does not affect the other occurring. For the two events to be independent then;
P(DnH) = P(D)P(H)
Given P(D) = 0.5 and P(H) = 0.3
P(D)P(H) = 0.5 × 0.3
P(D)P(H) = 0.15
And since P(DnH) = 0.2, hence P(DnH) ≠ P(D)P(H)
This means that the event "accepted at Harvard" IS NOT independent of the event "accepted at Dartmouth" since the two values are not equal.
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If f(x) = 2x – 5 and g(x) = x +4, then f(g(2)) =
Angle of the Sun A 96-ft tree casts a shadow that is 120 ft long. What is the angle of elevation of the sun
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.
Answers
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.
Answers
Firstly, we will find corner points
A=(-2,5)
B=(6,7)
C=(7,4)
the transformation that has the rule (x, y)→(−x, −y)
x---->-x
We will multiply x-values by -1
A=(-2*-1,5)=(2,5)
B=(6*-1,7)=(-6,7)
C=(7*-1,4)=(-7,4)
y---->-y
We will multiply y-values by -1
A=(2,5*-1)=(2,-5)
B=(-6,7*-1)=(-6,-7)
C=(-7,4*-1)=(-7,-4)
now, we can draw points and find graph
we get
Answers
solution:
a) To calculate the volume of cube of edge 1.0cm
Volume= (edge)³
=(100cm) ³
= 1cm³
To find the mass of MG atoms
Density = mass/volume
Mass of mg = 1.74g/cm³ x 1cm³
= 1.74g
To find the number of mg atoms,
Molar mass of mg = 24.31g
24.31g mg contain 6.022 x 10²³ atoms of mg
1.74g mg contain = 6.022 x 10²³mg atoms/24.31g mg x 1.74g mg
= 0.4310 x 10²³mg atoms
= 4.31 x 10²² mg atoms
B) to find the volume occupied by mg atoms
Total volume of cube = 1 cm³
Volume occupied by mg atoms
= 74% of 1cm³
=74/100 x 1cm³= 0.74cm³
To find the volume of 1 mg atom
4.31 x 10²² mg atoms occupy 0.74 cm³
1 atom will occupy 0.74cm³
1 atom will occupy = 0.74cm³/4.31 x 10²²mg atoms x 1mg atom
volume of 1 mg atom = 0.1716 x 10⁻²²cm²
Volume of 1 mg atom = 4/3 π³ in the form of sphere.
4/3πr³ = 0.1716 x 10⁻²²cm³
R³ = 0.1716 x 10⁻²²cm³x 4/3 x 7/22
R³ = 0.04095 x 10⁻²²cm³
R = 0.1599 x 10⁻⁷cm
1 = 10⁻¹⁰cm
= 0.1599 x 10⁻⁷cm
= 1pm/1 x 10⁻¹⁰cm x 0.1599 x 10-7cm
= 1.599 x 10² pm
R= 1.6 x 10²pm
symmetry for y=x²+2
Answers
Answer:
x=0
Step-by-step explanation:
The axis of symmetry always passes through the vertex of the parabola . The x -coordinate of the vertex is the equation of the axis of symmetry of the parabola. For a quadratic function in standard form, y=ax2+bx+c , the axis of symmetry is a vertical line x=−b/2a
in our case
y=x^2+2
a=1 b=0 c=2
so x=-b/2a=-0/2*1=0
x=0 is axis of symmetry
Answers
Answer:
y-3 = 2/9 (x-8)
Step-by-step explanation:
Coe-.00012t
where co is the initial anong . estimate the age of a sample of
wood discoverd by a arecheologist if the carbon level in the sample
is only 20% of it orginal carbon 14 level.
Answers
Answer:
The age of this sample is 13,417 years.
Step-by-step explanation:
The amount of carbon 14 present in a sample after t years is given by the following equation:
Estimate the age of a sample of wood discoverd by a arecheologist if the carbon level in the sampleis only 20% of it orginal carbon 14 level.
The problem asks us to find the value of t when
So:
The age of this sample is 13,417 years.