In a study of the progeny of rabbits, Fibonacci (ca. 1170-ca. 1240) encountered the sequence now bearing his name. The sequence
1 answer:
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Answer:
a. ;
b.
Step-by-step explanation:
We need here to apply the <em>Multiplication Principle </em>or the <em>Fundamental Principle of Counting</em> for each answer. Answer <em>b</em> needs an extra reasoning for being completed.
The <em>Multiplication Principle</em> states that if there are <em>n</em> ways of doing something and <em>m</em> ways of doing another thing, then there are <em>n</em> x <em>m</em> ways of doing both (<em>Rule of product</em> (2020), in Wikipedia).
<h3>In how many ways can ten students line up? </h3>There are <em>ten</em> students. When one is selected, there is no other way to select it again. So, <em>no repetition</em> is allowed.
Then, in the beginning, there are 10 possibilities for 10 students; when one is selected, there are nine possibilities left. When another is selected, eight possibilities are left to form the file, and so on.
Thus, we need to multiply the possibilities after each selection: that is <em>why</em> the <em>Multiplication Principle</em> is important here.
This could be expressed mathematically using n!:
For instance,
So, for the case in question, the <em>ten</em> students can line up in:
ways to line up in a single file.
For this question, we need to consider the former reasoning with extra consideration in mind.
The members of Group 1 can occupy <em>only</em> the following places in forming the file:
<em>places</em>.
The members of Group 2 <em>only</em>:
<em>places</em>.
And the members of Group 3, the following <em>only</em> ones:
<em>places.</em>
Well, having into account these possible places for each member of G1, G2 and G3, there are: <em>10! ways</em> for lining up members of G1; <em>10! ways</em> for lining up members of G2 and, also, <em>10! ways</em> for lining up members of G3.
After using the <em>Multiplication Principle</em>, we have, thus:
<em>ways the students can line up to come in from recess</em>.
Answer:
Step-by-step explanation:
Lets call x,y the numbers we obtain from the calculator. x and y are independent random variables of uniform[0,1] distribution.
Lets note that, since both x and y are between 0 and 1, then 1 is the biggest side of the triangle.
Lets first make a geometric interpretation. If the triangle were to be rectangle, then the side of lenght 1 should be its hypotenuse, and therefore x and y should satisfy this property:
x²+y² = 1
Remember that in this case we are supposing the triangle to be rectangle. But the exercise asks us to obtain an obtuse triangle. For that we will need to increase the angle obtained by the sides of lenght x and y. We can do that by 'expanding' the triangle, but if we do that preserving the values of x and y, then the side of lenght 1 should increase its lenght, which we dont want to. Thus, if we expand the triangle then we should also reduce the value of x and/or y so that the side of lenght 1 could preserve its lenght. With this intuition we could deduce that
x²+y² < 1
Now lets do a more mathematical approach. According to the Cosine theorem, a triangle of three sides of lenght a,b,c satisfies
a² = b²+c² - 2bc* cos(α), where α is the angle between the sides of lenght b and c.
Aplying this formula to our triangle, we have that
, where
is the angle between the sides of lenght x and y.
Since the triangle is obtuse, then , and for those values
is negative , hence we also obtain
1 > x² + y²
Thus, we need to calculate P(x²+y² <1). This probability can be calculated throught integration. We need to use polar coordinates.
(x, y) = (r*cosФ,r*senФ)
Where r is between 0 and 1, and Ф is between 0 and (that way, the numbers are positive).
The jacobian matrix has determinant r, therefore,
As a conclusion, the probability of obtaining an obtuse triangle is .