Given one factor of the polynomial, factor the polynomial completely.
1 answer:
You might be interested in
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 balls. We want to pick a white one, so we have a probability of
of picking a white one.
If this happens, we're left with white balls and still
black balls, for a total of
balls. So, now, the probability of picking a white ball is
The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability
Case 2: both balls are black
The exact same logic leads to a probability of
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
Case 1: both balls are white.
In this case, nothing changes between the two picks. So, you have a probability of 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
of picking a black ball with both picks.
This leads to an overall probability of
Of picking two balls of the same colour.
<h2>(c)</h2>We want to prove that
Expading all squares and products, this translates to
As you can see, this inequality comes in the form
With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:
And this is our case, because in our case we have
so, y has an extra piece and it is larger
which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2
(Простите, пожалуйста, мой английский. Русский не мой родной язык. Надеюсь, у вас есть способ перевести это решение. Если нет, возможно, прилагаемое изображение объяснит достаточно.)
Use the shell method. Each shell has a height of 3 - 3/4 <em>y</em> ², radius <em>y</em>, and thickness ∆<em>y</em>, thus contributing an area of 2<em>π</em> <em>y</em> (3 - 3/4 <em>y</em> ²). The total volume of the solid is going to be the sum of infinitely many such shells with 0 ≤ <em>y</em> ≤ 2, thus given by the integral
Or use the disk method. (In the attachment, assume the height is very small.) Each disk has a radius of √(4/3 <em>x</em>), thus contributing an area of <em>π</em> (√(4/3 <em>x</em>))² = 4<em>π</em>/3 <em>x</em>. The total volume of the solid is the sum of infinitely many such disks with 0 ≤ <em>x</em> ≤ 3, or by the integral
Using either method, the volume is 6<em>π</em> ≈ 18,85. I do not know why your textbook gives a solution of 90,43. Perhaps I've misunderstood what it is you're supposed to calculate? On the other hand, textbooks are known to have typographical errors from time to time...
