Let P2 denote the vector space of all polynomials with degree less than or equal to 2. (a) Show that B = {1 + x + x 2 , 1 + 2x −
x 2 , 1 − 2x − x 2} is a basis for P2. (b) Find the coord
1 answer:
Answer:
The answer is in the the attachment
Step-by-step explanation:
The question you posted was incomplete but i believe below is the complete question;
Let P2 denote the vector space of all polynomials with degree less than or equal to 2. (
a) Show that B = {1 + x + x 2 , 1 + 2x − x 2 , 1 − 2x − x 2} is a basis for P2.
(b) Find the coordinate vector of p(x) = 1 + 2x + 3x 2 relative to the basis B
The answer is explained in the attachments.
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First, complete the square in the equation for the second circle to determine its center and radius:
<em>x</em> ² - 10<em>x</em> + <em>y</em> ² = 0
<em>x</em> ² - 10<em>x</em> + 25 + <em>y </em>² = 25
(<em>x</em> - 5)² + <em>y</em> ² = 5²
So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.
Now convert each equation into polar coordinates, using
<em>x</em> = <em>r</em> cos(<em>θ</em>)
<em>y</em> = <em>r</em> sin(<em>θ</em>)
Then
<em>x</em> ² + <em>y</em> ² = 100 → <em>r </em>² = 100 → <em>r</em> = 10
<em>x</em> ² - 10<em>x</em> + <em>y</em> ² = 0 → <em>r </em>² - 10 <em>r</em> cos(<em>θ</em>) = 0 → <em>r</em> = 10 cos(<em>θ</em>)
<em>y</em> = 5 → <em>r</em> sin(<em>θ</em>) = 5 → <em>r</em> = 5 csc(<em>θ</em>)
See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to <em>y</em> = 5 at the point (0, 5).
Split up the region at 3 angles <em>θ</em>₁, <em>θ</em>₂, and <em>θ</em>₃, which denote the angles <em>θ</em> at which the curves intersect. They are
<em>θ</em>₁ = 0 … … … by solving 10 = 10 cos(<em>θ</em>)
<em>θ</em>₂ = <em>π</em>/6 … … by solving 10 = 5 csc(<em>θ</em>)
<em>θ</em>₃ = 5<em>π</em>/6 … the second solution to 10 = 5 csc(<em>θ</em>)
Then the area of the region is given by a sum of integrals:
To compute the integrals, use the following identities:
sin²(<em>θ</em>) = (1 - cos(2<em>θ</em>)) / 2
cos²(<em>θ</em>) = (1 + cos(2<em>θ</em>)) / 2
and recall that
d(cot(<em>θ</em>))/d<em>θ</em> = -csc²(<em>θ</em>)
You should end up with an area of
We can verify this geometrically:
• the area of the larger circle is 100<em>π</em>
• the area of the smaller circle is 25<em>π</em>
• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line <em>y</em> = 5, has area 100<em>π</em>/3 - 25√3
Hence the area of the region of interest is
100<em>π</em> - 25<em>π</em> - (100<em>π</em>/3 - 25√3) = 125<em>π</em>/3 + 25√3
as expected.
Answer:
P = 0.008908
Step-by-step explanation:
The complete question is:
The table below describes the smoking habits of a group of asthma sufferers
Nonsmokers Light Smoker Heavy smoker Total
Men 303 35 37 375
Women 413 31 45 489
Total 716 66 82 864
If two different people are randomly selected from the 864 subjects, find the probability that they are both heavy smokers.
The number of ways in which we can select x subjects from a group of n subject is given by the combination and it is calculated as:
Now, there are 82C2 ways to select subjects that are both heavy smokers. Because we are going to select 2 subjects from a group of 82 heavy smokers. So, it is calculated as:
At the same way, there are 864C2 ways to select 2 different people from the 864 subjects. It is equal to:
Then, the probability P that two different people from the 864 subjects are both heavy smokers is: