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Oliga [24]
2 years ago
15

Solve the inequality then graph. (you can graph on paper) d+4<-5

Mathematics
2 answers:
klasskru [66]2 years ago
8 0

Answer: i got 2 anwers 20, -9

i think -9

Step-by-step explanation: Subtract  9  from both sides

           d < -9   1.2  d  + 9.000  <  0

soldi70 [24.7K]2 years ago
4 0

Answer:

25

Step-by-step explanation:

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Find the size of angle XYZ.<br>Give your answer to 1 decimal place.<br>Z<br>13 cm<br>x<br>Y<br>4 cm​
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Answer:

<h2><em>7</em><em>2</em><em>.</em><em>9</em><em>°</em></h2>

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<em>tan \: y \:  =  \frac{13}{4}  \\ y =  {tan}^{ - 1}  ( \frac{13}{4} ) \\ y = 72.9</em>

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Could mud wrestling be the cause of a rash contracted by Washington State University students in the spring of 1992? Physicians
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Thank you for posting your question here at brainly. Feel free to ask more questions.   
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The best and most correct answer among the choices provided by the question is </span><span>A. Between 0.05 and 0.01 </span> .    
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An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem
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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:

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  1. x=b^2+w^2
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  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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