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aleksandr82 [10.1K]
3 years ago
5

Please help picture attached below. 100 POINTS!

Mathematics
2 answers:
Irina-Kira [14]3 years ago
8 0
Total number of marbles: 12; number of blue marble = 4

Then P(choosing one blue) = 4/13 = 1/3 (answer C)
Nina [5.8K]3 years ago
8 0

The answer is C for sure

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(a) Let R = {(a,b): a² + 3b <= 12, a, b € z+} be a relation defined on z+)
grin007 [14]

Answer:

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Step-by-step explanation:

The relation R is an equivalence if it is reflexive, symmetric and transitive.

The order to options required to show that R is an equivalence relation are;

((a, b), (a, b)) ∈ R since a·b = b·a

Therefore, R is reflexive

If ((a, b), (c, d)) ∈ R then a·d = b·c, which gives c·b = d·a, then ((c, d), (a, b)) ∈ R

Therefore, R is symmetric

If ((c, d), (e, f)) ∈ R, and ((a, b), (c, d)) ∈ R therefore, c·f = d·e, and a·d = b·c

Multiplying gives, a·f·c·d = b·e·c·d, which gives, a·f = b·e, then ((a, b), (e, f)) ∈R

Therefore R is transitive

From the above proofs, the relation R is reflexive, symmetric, and transitive, therefore, R is an equivalent relation.

Reasons:

Prove that the relation R is reflexive

Reflexive property is a property is the property that a number has a value that it posses (it is equal to itself)

The given relation is ((a, b), (c, d)) ∈ R if and only if a·d = b·c

By multiplication property of equality; a·b = b·a

Therefore;

((a, b), (a, b)) ∈ R

The relation, R, is reflexive.

Prove that the relation, R, is symmetric

Given that if ((a, b), (c, d)) ∈ R then we have, a·d = b·c

Therefore, c·b = d·a implies ((c, d), (a, b)) ∈ R

((a, b), (c, d)) and ((c, d), (a, b)) are symmetric.

Therefore, the relation, R, is symmetric.

Prove that R is transitive

Symbolically, transitive property is as follows; If x = y, and y = z, then x = z

From the given relation, ((a, b), (c, d)) ∈ R, then a·d = b·c

Therefore, ((c, d), (e, f)) ∈ R, then c·f = d·e

By multiplication, a·d × c·f = b·c × d·e

a·d·c·f = b·c·d·e

Therefore;

a·f·c·d = b·e·c·d

a·f = b·e

Which gives;

((a, b), (e, f)) ∈ R, therefore, the relation, R, is transitive.

Therefore;

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Based on a similar question posted online, it is required to rank the given options in the order to show that R is an equivalence relation.

Learn more about equivalent relations here:

brainly.com/question/1503196

4 0
3 years ago
Describe how a postulate becomes a theorem.
algol13
Postulate<span>: A statement, also known as </span>an axiom<span>, which is taken to be true without proof. Postulatesare the basic structure from which lemmas and </span>theorems<span> are derived. ... corollary: A proposition that follows with little or no proof required from one already proven( A deduction or an inference).</span>
8 0
3 years ago
The measure of angle MBT ?
Alik [6]

Answer:

Oooo I’m doing a question just like that one on my hw

Step-by-step explanation:

3 0
4 years ago
the amount tomas makes, in dollars, working h hours can be represented by the expression 18h+8. Tomas hopes to get a raise that
dangina [55]
20h+40 two h plus eighteen h is 20h, while eight plus 32 is 40. subsitute eight into it, u get 160, then add 40, which is 200
$200
8 0
4 years ago
The voltage across the capacitor increases as a function of time when an uncharged capacitor is placed in a single loop with a r
dmitriy555 [2]

Answer:

1. Exponential

Step-by-step explanation:

The simplest RC-Circuit, that is, a capacitor and a resistor in a series configuration can be modeled by using Ohm's Law and Kirchhoff's Circuit Laws:

C \cdot \frac{dV}{dt} + \frac{V}{R} = 0

By rearranging the formula, an homogeneous linear first-order differential equation is found:

\frac{dV}{dt} + \frac{1}{R \cdot C} \cdot V = 0

Whose solution has the form of a exponential model:

V(t) = V_{o} \cdot e^{-\frac{t}{R \cdot C} }

7 0
3 years ago
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