Let A and B be events defined on a sample space S. If the probabilit.pdf
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Let A and B be events defined on a sample space S. If the probability that at least one of them occurs is 0.3 and the probability that A occurs but B does not is 0.1, what is P(B)? Solution P(AUB)=0.3 P(AUB\')=0.1 or P(B\')-P(A\'B\')=0.1 P(AUB)=P(A)+P(B)-P(AB) P(A\'B\')=1-P(AUB) =0.7 or 1-P(B)=0.8 or P(B)=0.2.
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Let F(x,y) be the statement x can fool y,where the domainconsi.pdf
Let F(x,y) be the statement \"x can fool y,\"where the domainconsists of all people in the world.
Use quantifiers to expresseach of these statements.
1) Everybody can fool fred.
2) Evelyn can fool everybody.
3) Everybody can fool somebody.
4) There is no one who can fool everybody
5) Everyone can be fooled by somebody.
6) No one can fool both fred and jerry.
7) Nancy can fool exactly two people.
8) There is exactly one person whom everybody can fool.
9) No one can fool himself or herself.
10)There is someone who can fool exactly one person besideshimself or herself.
Let F(x,y) be the statement \"x can fool y,\"where the domainconsists of all people in the world.
Use quantifiers to expresseach of these statements.
1) Everybody can fool fred.
2) Evelyn can fool everybody.
3) Everybody can fool somebody.
4) There is no one who can fool everybody
5) Everyone can be fooled by somebody.
6) No one can fool both fred and jerry.
7) Nancy can fool exactly two people.
8) There is exactly one person whom everybody can fool.
9) No one can fool himself or herself.
10)There is someone who can fool exactly one person besideshimself or herself.
Solution
1. for all x, F(x, fred) 2. for all x, F(evelyn, x) 3. for all x, there exists y, F(x, y) 4
not there exists y, for all x, F(y, x) [or,equivalently, for all y, there exists x, not F(y, x)] 5 for all
x, there exists y, F(y, x) 6 not there exists x, F(x, fred) and F(x, jerry) 7 (there exists x, there
exists y, x not equal y andF(nancy, x) and F(nancy, y)) and not (there exists x, there existsy,
there exists z, x not equal y, x not equal z, y not equal z, andF(nancy, x) and F(nancy, y) and
F(nancy, z)) 8 (there exists x, for all y F(y, x)) and not (thereexists x, there exists y, x not equal
y, and for all z F(z, x) andF(z, y)) 9 not there exists x, F(x, x) 10 there exists x, there exists y, x
not equal y and F(x, y)and for all z not equal y, x, not F(x, z).
Let F(x, y) be the statement x can fool y, where the universe of.pdf
Let F(x, y) be the statement \"x can fool y,\" where the universe of discourse consists of all the
human beings on Earth. Translate the statement into plain English:
a. There is someone that can fool everyone.
b. There is one person who can be fooled by everybody.
c. Everyone can fool somebody.
d. Everyone can be fooled by somebody.
Solution
it says for any x, x can fool y => this means that every person can fool atleast some
person. therefore (c).
let f(x) = csc^2xfind f(x) find f(4)SolutionTreat csc.pdf
let f(x) = csc^2x
find f\'(x)
find f\'(/4)
Solution
Treat csc x just like any other variable squared (dy/dx x^2 = 2x)
So dy/dx csc^2(x) is 2csc(x)
but now you have to take the derivative of the \"inside\" function, which is csc (x):
dy/dx csc(x) = -csc(x)cot(x)
multiply the two and you get your answer:
so f\'(x) = dy/dx (csc^2(x)) = -2csc^2(x)cot(x)
so f\'(pi/4)=-2*(2)(1)=-4.
Let f(x) and g(x) be increasing and Riemann integrable. h is defined.pdf
Let f(x) and g(x) be increasing and Riemann integrable. h is defined as h=f-g. Show that h is
Riemann integrable.
Solution
f(x) is increasing and Riemann integrable.
g(x) is also increasing and Riemann integrable.
As f(x) and g(x) are both Riemann integrable, we can say the limit of the sum of the partitioned
area exists.
This again implies the difference of the function also is Riemann integrable.
As f(x) - g(x) also will have limit as L1-L2 where L1 is Riemann limit of f(x) and L2 is Riemann
limit of g(x)
Thus h(x) = f(x) - g(x) also will have a limit L1-L2
Hence h(x) is Riemann integrable..
Let f be the p.d.f of a continuous random variables. Is it always tr.pdf
Let f be the p.d.f of a continuous random variables. Is it always true that f(x) less and equal to 1
for all x in the range of X? Does it have to be true for some x in the range of X??
Solution
Is it always true that f(x) less and equal to 1 for all x in the range of X? No,
f(x) is not a probability (like the cdf). f(x) can be bigger than 1. For example, for uniform
distribution on [0,1/2], f(x)=2 Does it have to be true for some x in the range of X?? Yes,
otheriwse if f(x)>1 always then 1=integral f(x) >1 contradiction.
Let f be continuous on the interval [0, 1] to R and such that f(0)=f.pdf
Let f be continuous on the interval [0, 1] to R and such that f(0)=f(1). Prove that there exists a
point c in [0, 1/2] such that f(c) = f(c + 1/2).
Solution
This is actually Rolles theorm which is true for all numbers that satisfy the given
condition.
Let F be a field. Prove that the only ideals of F are F and {0}S.pdf
Let F be a field. Prove that the only ideals of F are F and {0}
Solution
It should be clear that {0} is an ideal of F. Now, suppose I is an ideal of F containing some
nonzero element x. Then x is a unit, so it has an inverse, call it x^-1. Then, since I is an ideal, it
contains x(x^-1) = 1. Now, let y ? F. Since I is an ideal, we have y = 1(y) ? I. Thus, F = I..
Let c represent I was my care, let r represent it rains and .pdf
Let c represent \"I was my care,\" let r represent \"it rains\" and let p represent \"the play is
postponed\" Write the compound statement in symbols.
If it does not rain, then I was my car.
The statement is written symbolically as ~____---->_____
Solution
p and -p.
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Let F(x,y) be the statement x can fool y,where the domainconsi.pdf
Let F(x,y) be the statement \"x can fool y,\"where the domainconsists of all people in the world.
Use quantifiers to expresseach of these statements.
1) Everybody can fool fred.
2) Evelyn can fool everybody.
3) Everybody can fool somebody.
4) There is no one who can fool everybody
5) Everyone can be fooled by somebody.
6) No one can fool both fred and jerry.
7) Nancy can fool exactly two people.
8) There is exactly one person whom everybody can fool.
9) No one can fool himself or herself.
10)There is someone who can fool exactly one person besideshimself or herself.
Let F(x,y) be the statement \"x can fool y,\"where the domainconsists of all people in the world.
Use quantifiers to expresseach of these statements.
1) Everybody can fool fred.
2) Evelyn can fool everybody.
3) Everybody can fool somebody.
4) There is no one who can fool everybody
5) Everyone can be fooled by somebody.
6) No one can fool both fred and jerry.
7) Nancy can fool exactly two people.
8) There is exactly one person whom everybody can fool.
9) No one can fool himself or herself.
10)There is someone who can fool exactly one person besideshimself or herself.
Solution
1. for all x, F(x, fred) 2. for all x, F(evelyn, x) 3. for all x, there exists y, F(x, y) 4
not there exists y, for all x, F(y, x) [or,equivalently, for all y, there exists x, not F(y, x)] 5 for all
x, there exists y, F(y, x) 6 not there exists x, F(x, fred) and F(x, jerry) 7 (there exists x, there
exists y, x not equal y andF(nancy, x) and F(nancy, y)) and not (there exists x, there existsy,
there exists z, x not equal y, x not equal z, y not equal z, andF(nancy, x) and F(nancy, y) and
F(nancy, z)) 8 (there exists x, for all y F(y, x)) and not (thereexists x, there exists y, x not equal
y, and for all z F(z, x) andF(z, y)) 9 not there exists x, F(x, x) 10 there exists x, there exists y, x
not equal y and F(x, y)and for all z not equal y, x, not F(x, z).
Let F(x, y) be the statement x can fool y, where the universe of.pdf
Let F(x, y) be the statement \"x can fool y,\" where the universe of discourse consists of all the
human beings on Earth. Translate the statement into plain English:
a. There is someone that can fool everyone.
b. There is one person who can be fooled by everybody.
c. Everyone can fool somebody.
d. Everyone can be fooled by somebody.
Solution
it says for any x, x can fool y => this means that every person can fool atleast some
person. therefore (c).
let f(x) = csc^2xfind f(x) find f(4)SolutionTreat csc.pdf
let f(x) = csc^2x
find f\'(x)
find f\'(/4)
Solution
Treat csc x just like any other variable squared (dy/dx x^2 = 2x)
So dy/dx csc^2(x) is 2csc(x)
but now you have to take the derivative of the \"inside\" function, which is csc (x):
dy/dx csc(x) = -csc(x)cot(x)
multiply the two and you get your answer:
so f\'(x) = dy/dx (csc^2(x)) = -2csc^2(x)cot(x)
so f\'(pi/4)=-2*(2)(1)=-4.
Let f(x) and g(x) be increasing and Riemann integrable. h is defined.pdf
Let f(x) and g(x) be increasing and Riemann integrable. h is defined as h=f-g. Show that h is
Riemann integrable.
Solution
f(x) is increasing and Riemann integrable.
g(x) is also increasing and Riemann integrable.
As f(x) and g(x) are both Riemann integrable, we can say the limit of the sum of the partitioned
area exists.
This again implies the difference of the function also is Riemann integrable.
As f(x) - g(x) also will have limit as L1-L2 where L1 is Riemann limit of f(x) and L2 is Riemann
limit of g(x)
Thus h(x) = f(x) - g(x) also will have a limit L1-L2
Hence h(x) is Riemann integrable..
Let f be the p.d.f of a continuous random variables. Is it always tr.pdf
Let f be the p.d.f of a continuous random variables. Is it always true that f(x) less and equal to 1
for all x in the range of X? Does it have to be true for some x in the range of X??
Solution
Is it always true that f(x) less and equal to 1 for all x in the range of X? No,
f(x) is not a probability (like the cdf). f(x) can be bigger than 1. For example, for uniform
distribution on [0,1/2], f(x)=2 Does it have to be true for some x in the range of X?? Yes,
otheriwse if f(x)>1 always then 1=integral f(x) >1 contradiction.
Let f be continuous on the interval [0, 1] to R and such that f(0)=f.pdf
Let f be continuous on the interval [0, 1] to R and such that f(0)=f(1). Prove that there exists a
point c in [0, 1/2] such that f(c) = f(c + 1/2).
Solution
This is actually Rolles theorm which is true for all numbers that satisfy the given
condition.
Let F be a field. Prove that the only ideals of F are F and {0}S.pdf
Let F be a field. Prove that the only ideals of F are F and {0}
Solution
It should be clear that {0} is an ideal of F. Now, suppose I is an ideal of F containing some
nonzero element x. Then x is a unit, so it has an inverse, call it x^-1. Then, since I is an ideal, it
contains x(x^-1) = 1. Now, let y ? F. Since I is an ideal, we have y = 1(y) ? I. Thus, F = I..
Let c represent I was my care, let r represent it rains and .pdf
Let c represent \"I was my care,\" let r represent \"it rains\" and let p represent \"the play is
postponed\" Write the compound statement in symbols.
If it does not rain, then I was my car.
The statement is written symbolically as ~____---->_____
Solution
p and -p.
Let B= { 1+x, x+x^2, 1+x^2} be an ordered basis of P2. And let B b.pdf
Let B= { 1+x, x+x^2, 1+x^2} be an ordered basis of P2. And let B\' be the standard basis of P2.
Find the coordinate vector of p(x) = 1+2x-x^2 with respect to B. Please show steps.
Solution
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2+2[1+x]i
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Let E and F be two events for which one knows that the probability t.pdf
Let E and F be two events for which one knows that the probability that atleast one of them
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Solution
P(A or B) = P(A) + P(B) - P(A & B) = .21+.34-.05 = .5 P(neither A nor B) = 1 -
P(A or B) = .5.
Let A= {-1, 0, 1}. Which of the following field properties does not .pdf
Let A= {-1, 0, 1}. Which of the following field properties does not hold for this set?
Identity for addition
Commutative for multiplication
Inverse for multiplication
Distributive
Identity for addition
Commutative for multiplication
Inverse for multiplication
Distributive
Solution
Inverse for multiplication
remaining all other option the set holds
you can verify it bi dividing the A in to 2 or more subsets
thank you.
Let A be an nSolutionAs there are two eigen values lemda 1 and.pdf
Let A be an n
Solution
As there are two eigen values lemda 1 and lemda 2 we must get n linearly independent eigen
vectors to have A diagonalizable
As for lemda 1 we have n-1 dimensional null space we have n-1 eigen vectors.
For lemda 2 atleast one eigen vector we have
together we have n eigen distinct vectors
Hence A is diagonalisable.
Let A be a set of n distinct elements. then the total number ofdisti.pdf
Let A be a set of n distinct elements. then the total number ofdistinct functions from A to A is
and out of these are onto functions
Solution
Each input has n possible outputs, and there are a total of ninputs so, the number of distinct
functions from A to A isnn.
The first input has n possible outputs, the next input has n-1possible outputs (all outputs
exculding the first selected), thenext has n-2 (all outputs excluding the first two
selected),continuing on in this vein, it can be concluded, the number ofdistinct onto functions is
n!.
Let A = [ 2 1 0 3][ 4 4 1 5][ 4 2 3 1]Find the LU factorizatio.pdf
Let A = [ 2 1 0 3]
[ 4 4 1 5]
[ 4 2 3 1]
Find the LU factorization for A. I have found U but I am not sure how to get L. Show me how to
find L.
Solution
For the given matrix LU decomposition is not possible because the given matrix is recangular.
Let A and B be events defined on a sample space S. Suppose that P(A).pdf
Let A and B be events defined on a sample space S. Suppose that P(A) = 0.4, P(B) = 0.5, and the
probability of both occurring is 0.1. What is the probability that either A or B (but not both)
occur?
Solution
P(A) = 0.4 P(B) = 0.5 P(A and B) = 0.1 P(A or B) = P(A) + P(B) - P(A and B) =
0.4 + 0.5 - 0.1 = 0.8 P(either A or B not both) = P(A or B) - P(A and B) = 0.8 - 0.1 = 0.7.
Let A = IN, the set of all positive integers. In each case, give an.pdf
Let A = IN, the set of all positive integers. In each case, give an
example of a function f : A -> A with the indicated properties, or
explain why no such function exists.
(a) f is not the identity function on A, but f is bijective.
(b) f is one-to-one, but not onto
Solution
as it form one to many relation ship . so it is not a function..
Let A and define f A by f(x) = ceil(x). Define the set A to.pdf
Let A ? ? and define f: A ? ? by f(x) = ceil(x). Define the set A to ensure that f is injective on A.
Solution
since range of ceil function is all the integers.
so for f to be injective, f should be one-one.
and f to be one-one there should not exist two elements in domain that is in A such that when we
apply f on that elements output is same.
and it will be possible only when
A = (set of all integers).
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Solution
P(A) = 0.4 P(B) = 0.5 P(A and B) = 0.1 P(A or B) = P(A) + P(B) - P(A and B) =
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(b) f is one-to-one, but not onto
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