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ICE Princess25 [194]
3 years ago
8

A line passes through the points(-9,9) and has a slope of 2/3​

Mathematics
1 answer:
Anni [7]3 years ago
4 0
ANSWER

y = 2/3x + 15

Explanation:

Equation used: y - y1 = m( x - x1 )

y - 9 = 2/3 ( x - -9 )

y - 9 = 2/3x + 6 ( distribution )

y - 9 = 2/3x + 6 ( add 9 to 6 )

y = 2/3x + 15
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The quotient is -2.


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Which shows the correct substitution of the values a, b, and c from the equation 1 = -2x + 3x2 + 1 into the quadratic
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Answer: A

Step-by-step explanation:

The first thing we need to do is to make sure our equation is in standard form. The given equation is not in standard form.

3x²-2x=0

Now that the equation is in standard form, we can find our A, B, C to see which quadratic equation is correct.

A=3

B=-2

C=0

We can automatically eliminate D because the first value is -3 when it is supposed to be -(-2). Also, the denominator is supposed to be 2a. We know that A=3. The denominator should be 2(3), not 2(-2).

Next, we can eliminate B and C. For the 4ac, we know it has to be 4(3)(0) because A=3 and C=0. B and C have given 4(3)(2) and 4(3)(1), respectively. This does not match A=3 and C=0.

Therefore, A is the correct answer.

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Solve for x. Round your answer to the nearest tenth if necessary.
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Step-by-step explanation:

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2 years ago
An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem
Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

Case 1: both balls are white.

At the beginning we have b+w balls. We want to pick a white one, so we have a probability of \frac{w}{b+w} of picking a white one.

If this happens, we're left with w-1 white balls and still b black balls, for a total of b+w-1 balls. So, now, the probability of picking a white ball is

\dfrac{w-1}{b+w-1}

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}

Case 2: both balls are black

The exact same logic leads to a probability of

\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of \frac{w}{b+w} of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability \frac{b}{b+w} of picking a black ball with both picks.

This leads to an overall probability of

\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}

Of picking two balls of the same colour.

<h2>(c)</h2>

We want to prove that

\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

Expading all squares and products, this translates to

\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}

As you can see, this inequality comes in the form

\dfrac{x}{y}\geq \dfrac{x-k}{y-k}

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y

And this is our case, because in our case we have

  1. x=b^2+w^2
  2. y=b^2+w^2+2bw so, y has an extra piece and it is larger
  3. k=b+w which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

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