Suppose that B1 and B2 are mutually exclusive and complementary events, such that P(B1 ) = .6 and P(B2) = .4. Consider another event A such that P(A | B1) = .2 and P(A | B2) = .5. Find P(A).
Answers
Answer:
So, we get that is P(A)=0.32.
Step-by-step explanation:
We know that:
We have the formula for probability:
So, we calculate:
We calculate:
So, we get that is P(A)=0.32.
Final answer:
To find P(A), use the law of total probability given that B1 and B2 are mutually exclusive and complementary events. Substituting the provided values, P(A) = 0.32.
Explanation:
The question is asking us to calculate P(A), given the values for P(A | B1) and P(A | B2), and the knowledge that B1 and B2 are mutually exclusive and complementary events. In probability, if events B1 and B2 are mutually exclusive and complementary, this means that one and only one of them can occur, and their occurrence covers all possible outcomes. We can use the law of total probability to find the overall P(A). The law of total probability states that P(A) = P(A | B1) * P(B1) + P(A | B2) * P(B2). Plugging the provided values into this formula, we get P(A) = .2 * .6 + .5 * .4 = .12 + .2 = .32. Therefore, P(A) is .32.
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A college professor states that this year's entering students appear to be smarter than entering students from previous years. The college's records indicate that the mean IQ for entering students from earlier years. is 110.3. Suppose we want to sample a smaller number of this year's entering students and carry out a hypothesis test to see if the professor statement can be supported. State the null and alternative hypothesis.
O -$3.008.23
O $608.23
$591.77
O $1,808.23
$3.008.23
Intro
Done
Answers
Answer:
591.77
Step-by-step explanation:
we multiply 600 by 3 to get 1,800 then subtract 1,208.23 from 1,800 to get 591.77
8
of her collection.
5
7
of the remaining books are Chinese books and the rest are German books. How many books are there in her book collection if there are 75 Chinese books?
Answers
There are 168 books are there in her book collection if there are 75 Chinese books.
Given
Miko has a vast collection of books in her library.
English books = 3/8
Chinese books = 5/7.
Ratio proportion;
A ratio is a comparison of two or more numbers that indicates their quantities in relation to each other.
The proportion of the English books is 3/8, the remaining is:
Then,
Chines books = ratio of Chinese book × ratio of remaining books × Total books
Hence, there are 168 books are there in her book collection if there are 75 Chinese books.
To know more about proportion click the link given below.
Answer:
168 books
Step-by-step explanation:
Given
of the remaining books
Required
Determine the total number of books
If the proportion of the English books is 3/8, the the remaining is:
Take LCM
From the question, we understand that:
of the remaining books
This means:
From the question, we also understand that 75 are Chinese books.
The total number of books can then be calculated using:
Substitute 75 for Chinese
Make Total the subject:
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.
Solve the formula for b.
Answers
Answer:
(2A)/h -a =b
Step-by-step explanation:
A= [h(a+b)]/2 I rewrote the formula, because h and a+b are in the numerator.
2A=h(a+b) Multiply both sides of equation by 2
(2A)/h= a+b Divide both side by h
(2A)/h -a= b Subtract a from both sides
Answers
Answer:
-690/1000
Step-by-step explanation:
B. 9 - 2
C. y/9 = 3
D. 7 + y