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3 years ago

Which sets of values belong to the domain of range of a relation?

Mathematics

2 answers:

aliina [53]3 years ago

5 0

Domain is all X values, and range is the Y values.

Goryan [66]3 years ago

4 0

For Domain:

values for the independent variable

input values

For range:

values for the dependent variable

output values

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(0, 10) and (24, 6)

Inessa05 [86]

As in which form do u want that in?

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3 years ago

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Which Best summarizes the Pythagorean theorem

Masja [62]

the sum of the squares of the lengths of the legs of a right triangle equal the square of the length of the hypotenuse

Explanation

Pythagorean theorem is related to the right angle triangle and it states that

c^2= a^2 + b^2

Where C is the length of the hypotenuse

a and b are the length of the sides of the triangle

The theorem states that the sum of the squares of the lengths of the sides of a right triangle is equal to the square of the length of the hypotenuse

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3 years ago

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Lin bought x grams of sugar. Jada bought 3⁄4 more than that. what is the equation that will represents the relationship between

astra-53 [7]

Answer:

y = 7/4x

Step-by-step explanation:

Lin bought x grams of sugar.

Jada bought 3⁄4 more than Lin bought.

<u>The equation:</u>

y = x + 3/4x ⇒ y = 7/4x

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2 years ago

For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.

nordsb [41]

Answer:

<h3>For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.</h3>

By De morgan's law

$(A\cap B)^{c}=A^{c}\cup B^{c}\\\\P((A\cap B)^{c})=P(A^{c}\cup B^{c})\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq 1-P(A)+1-P(B)\\\\-P(A\cap B)\leq 1-P(A)-P(B)\\\\P(A\cap B)\geq P(A)+P(B)-1$

which is Bonferroni’s inequality

<h3>Result 1: P (Ac) = 1 − P(A)</h3>

Proof

If S is universal set then

$A\cup A^{c}=S\\\\P(A\cup A^{c})=P(S)\\\\P(A)+P(A^{c})=1\\\\P(A^{c})=1-P(A)$

<h3>Result 2 : For any two events A and B, P (A∪B) = P (A)+P (B)−P (A∩B) and P(A) ≥ P(B)</h3>

Proof:

If S is a universal set then:

$A\cup(B\cap A^{c})=(A\cup B) \cap (A\cup A^{c})\\=(A\cup B) \cap S\\A\cup(B\cap A^{c})=(A\cup B)$

Which show A∪B can be expressed as union of two disjoint sets.

If A and (B∩Ac) are two disjoint sets then

$P(A\cup B) =P(A) + P(B\cap A^{c})---(1)\\$

B can be expressed as:

$B=B\cap(A\cup A^{c})\\$

If B is intersection of two disjoint sets then

$P(B)=P(B\cap(A)+P(B\cup A^{c})\\P(B\cup A^{c}=P(B)-P(B\cap A)$

Then (1) becomes

$P(A\cup B) =P(A) +P(B)-P(A\cap B)\\$

<h3>Result 3: For any two events A and B, P(A) = P(A ∩ B) + P (A ∩ Bc)</h3>

Proof:

If A and B are two disjoint sets then

$A=A\cap(B\cup B^{c})\\A=(A\cap B) \cup (A\cap B^{c})\\P(A)=P(A\cap B) + P(A\cap B^{c})\\$

<h3>Result 4: If B ⊂ A, then A∩B = B. Therefore P (A)−P (B) = P (A ∩ Bc) </h3>

Proof:

If B is subset of A then all elements of B lie in A so A ∩ B =B

$A =(A \cap B)\cup (A\cap B^{c}) = B \cup ( A\cap B^{c})$

where A and A ∩ Bc are disjoint.

$P(A)=P(B\cup ( A\cap B^{c}))\\\\P(A)=P(B)+P( A\cap B^{c})$

From axiom P(E)≥0

$P( A\cap B^{c})\geq 0\\\\P(A)-P(B)=P( A\cap B^{c})\\P(A)=P(B)+P(A\cap B^{c})\geq P(B)$

Therefore,

P(A)≥P(B)

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3 years ago

What is the answer ‍♀️

N76 [4]

Answer:

c. 15in

Step-by-step explanation:

5 0

2 years ago

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