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m_a_m_a [10]
3 years ago
14

Factor -2.9 out of 8.7b – 11.6.

Mathematics
1 answer:
Mama L [17]3 years ago
6 0

Step-by-step explanation:

Since 8.7b = (-2.9) * (-3b),

and -11.6 = (-2.9) * (4),

8.7b - 11.6 = -2.9(4 - 3b).

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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
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
Reggie lost 3 spaceships in level 3 of a video game. He lost 45 points for each spaceship. When he completed level 3, he earned
Georgia [21]

Answer:

Reggie has 5 cockldos

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
O panglica a fost taiata, prin 5 taieturi , in bucati cu lungimea de 35 cm . Cati metri masura panglica?
Amiraneli [1.4K]
35×5=175 (metri măsoară pamblica)

3 0
3 years ago
HELP ME THANKS +×+
Tresset [83]

Answer:

hypotenuse =  \sqrt{ {height}^{2} +  {adjacent}^{2}  }  \\  =  \sqrt{ {11}^{2}  +  {60}^{2} }  \\  =  \sqrt{3721}  \\  = 61 \: inches

4 0
3 years ago
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The energy released by metabolism of 1 average candy bar is 1 x 10^6 joules.
Elis [28]

Answer: 32

Step-by-step explanation:

Given: The energy needed for 1 hour of running for an adult =4.5\times10^6\ joules.

Therefore. the energy needed for seven hours of running==7\times4.5\times10^6\ joules.=31.5\times10^6\ joules

Since, he energy released by metabolism of 1 average candy bar =1\times10^6\ joules.

Therefore, the number of candy bars an adult need to eat  to supply the energy needed for seven hours of running.=\frac{31.5\times10^6}{1\times10^6}=31.5\approx32

Hence, the adult need to eat 32 candies to supply the energy needed for seven hours of running.

3 0
3 years ago
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