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3 years ago

Does choosing a blue marble represent the complement event of choosing a red marble? Explain.

Mathematics

3 answers:

Vaselesa [24]3 years ago

4 0

Answer:

No, the probability of choosing a red marble is not the complement of choosing a blue marble.

P(red) + P(blue) = 3/5, not 1.

The number of blue and red marbles does not add to the total number of outcomes.

There are other marbles besides red ones and blue ones.

Step-by-step explanation:

i got it right.

Arada [10]3 years ago

3 0

Answer:

Below is the sample response

Step-by-step explanation:

The complementary event for choosing a red marble is choosing a blue or purple. The total number of marbles is 5. The probability of choosing a red marble is 1/5, and the complement would be 4/5. That is not the same as choosing a blue marble.

Lina_smartkid2 years ago

0 0

Lina_smartkid

2 years ago

got it right on edg

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Answer:

case a) $x^{2}=3y$ ----> open up

case b) $x^{2}=-10y$ ----> open down

case c) $y^{2}=-2x$ ----> open left

case d) $y^{2}=6x$ ----> open right

Step-by-step explanation:

we know that

1) The general equation of a vertical parabola is equal to

$y=a(x-h)^{2}+k$

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open upward and the vertex is a minimum

If a<0 ----> the parabola open downward and the vertex is a maximum

2) The general equation of a horizontal parabola is equal to

$x=a(y-k)^{2}+h$

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open to the right

If a<0 ----> the parabola open to the left

Verify each case

case a) we have

$x^{2}=3y$

$y=(1/3)x^{2}$

$a=(1/3)$

$a>0$

therefore

The parabola open up

case b) we have

$x^{2}=-10y$

$y=-(1/10)x^{2}$

$a=-(1/10)$

$a$

therefore

The parabola open down

case c) we have

$y^{2}=-2x$

$x=-(1/2)y^{2}$

$a=-(1/2)$

$a$

therefore

The parabola open to the left

case d) we have

$y^{2}=6x$

$x=(1/6)y^{2}$

$a=(1/6)$

$a>0$

therefore

The parabola open to the right

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3 years ago