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timofeeve [1]
2 years ago
6

How do I do this? I’m not sure can anyone help?

Mathematics
2 answers:
eduard2 years ago
8 0

value of x= 7√3

Hope it helps you...

kiruha [24]2 years ago
6 0
Maybe same answer is correct
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What is 4/11 in decimal form?
Rudiy27
4/11 is 0.363636363636
And "36" just keeps repeating
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3 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

4 0
3 years ago
Recopier et compléter chaque phrase.
MrRissso [65]
Unnoticed 507 is:67 is 9$
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3 years ago
Beth and Bonnie bought a submarine sandwich. Beth cut the sandwich into four equal parts. Each girl ate half of her sandwich. Ho
padilas [110]
Each girl ate 1/4 hope this helps
6 0
3 years ago
Read 2 more answers
What is this please help?
mestny [16]

Answer:

y = 2/3x - 1

Step-by-step explanation:

The simpliest way to determine the slope of the line is to see the rise over run of the line. In this case, we are given two points which are at (0, -1) and (3,1)

By calculating rise of run, we can see that the starting point goes up by 2 units, from -1 to 1 and goes to the right 3 units, from 0 to 3

In this case, that means we have a slope of 2/3x.

The +b plays into the y-intercept of the line, which is actually the same of one of our points: (0,-1)

Since this number is a negative, it would actually be y = (2/3)x + (-1) which is the same as y=2/3x - 1

7 0
3 years ago
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