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3 years ago

Find the Probability of pulling a green marble from a bag with six green and eight blue marbles.

Mathematics

2 answers:

jeka57 [31]3 years ago

4 0

Answer:

6/14 or in its simplest form 3/7

frosja888 [35]3 years ago

3 0

3/7 is the right answer, just took the test on plato

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26. Define a relation ∼ ∼ on R 2 R2 by stating that ( a , b ) ∼ ( c , d ) (a,b)∼(c,d) if and only if a 2 + b 2 ≤ c 2 + d 2 . a2+

Tresset [83]

Answer:

~ is reflexive.

~ is asymmetric.

~ is transitive.

Step-by-step explanation:

~ is reflexive:

i.e., to prove $\forall (a, b) \in \mathbb{R}^2$ , $(a, b) R(a, b)$ .

That is, every element in the domain is related to itself.

The given relation is $\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$

Reflexive:

$(a, b) \sim (a, b)$ since $a^2 + b^2 = a^2 + b^2$

This is true for any pair of numbers in $\mathbb{R}^2$ . So, $\sim$ is reflexive.

Symmetry:

$\sim$ is symmetry iff whenever $(a, b) \sim (c, d)$ then $(c, d) \sim (a, b)$ .

Consider the following counter - example.

Let (a, b) = (2, 3) and (c, d) = (6, 3)

$a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13$

$c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42$

Hence, $(a, b) \sim (c, d)$ since $a^2 + b^2 \leq c^2 + d^2$

Note that $c^2 + d^2 \nleq a^2 + b^2$

Hence, the given relation is not symmetric.

Transitive:

$\sim$ is transitive iff whenever $(a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f)$ then $(a, b) \sim (e, f)$

To prove transitivity let us assume $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$ .

We have to show $(a, b) \sim (e, f)$

Since $(a, b) \sim (c, d)$ we have: $a^2 + b^2 \leq c^2 + d^2$

Since $(c, d) \sim (e, f)$ we have: $c^2 + d^2 \leq e^2 + f^2$

Combining both the inequalities we get:

$a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2$

Therefore, we get: $a^2 + b^2 \leq e^2 + f^2$

Therefore, $\sim$ is transitive.

Hence, proved.

3 0

3 years ago

1/3x - 3/2y = 4

Alisiya [41]

Multiply the first fraction by six and get : 2x - 9y = 24
So if this system has no solutions, then b would be nine.

3 0

3 years ago

What is the area of a sector with radius 6" and measure of arc equal to 120°?

fenix001 [56]

The area of a sector = 0.5 * radius^2 * central angle measure (in radians)

So that:

$Area_{sector}= \frac{1}{2} * 6^2* \frac{120}{180}*\pi=12\pi=37.69911
\\
Area_{sector}=37.7$ sq in

7 0

3 years ago

Represent the ratio 2:3 in as many ways possible

Ksju [112]

Answer:

HELP PLEASE

Step-by-step explanation:

Which of the following equations has an axis of symmetry with the equation x=1.5?

A. y=2x2−6x−1

B. y=2x2−3x+1

C. y=x2+6x−3

D. y=2x2+3x+2

7 0

3 years ago

Read 2 more answers

How do I get this graph and how do I get it with multiplicity polynomials

olganol [36]

1.If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero.

2.If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity. is the answer to the question

6 0

2 years ago