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astra-53 [7]
3 years ago
13

Find the solution for h - 12 =6

Mathematics
2 answers:
Ne4ueva [31]3 years ago
4 0

Answer:

H-12=6

H−12+12=6+12

h=18

likoan [24]3 years ago
4 0

Answer:

h=18

Step-by-step explanation:

h-12=6

is the normal eqaution, you need to determine the value of h, on the left side of the eqaution you have negative twelve we must get the h on its own by moving the 12 to the right side using of basic math rules you must know that when you have a negative and want to move it to the right hand side it must become a positive

so in formula it wil look like this

h = 6 + 12

Add up 6 with 12 that wil give you 18

so the value of h is then 18

h = 18

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To factorise this quadratic expression we'll find 2  factors of 5 that when added gives the coefficient of x which is

Factors = +1 and +5

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Rewriting equation

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x ( x +1) +5 ( x + 1)

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Answer:

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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

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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}

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<h2>(c)</h2>

We want to prove that

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\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

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