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ss7ja [257]
3 years ago
11

What is an a priori proposition? an a posteriori proposition?

Mathematics
1 answer:
solniwko [45]3 years ago
4 0
A priori means as things were in the beginning or at the start.
a posteriori means as things are after any event.
You might be interested in
Select the statement that correctly describes the solution to this system of equations.
jasenka [17]
Selection C is appropriate.

7 0
3 years ago
The volleyball team and the wrestling team at Stamford High School are having a joint car wash today,
Sedbober [7]

Answer:

Money raised by each team = 94 + 11 = <em>$105</em>

Number of cars washed = <em>11</em>

Step-by-step explanation:

Money already present with volleyball team = $50

Money raised by washing each car by volleyball team = $5

Let the number of cars washed = x

Money raised by washing cars = $5x

Money already present with volleyball team = $94

Money raised by washing each car by volleyball team = $1

Money raised by washing cars = $1x = $x

Given that, both the teams have raised same amount of money:

50+5x=94+x\\\Rightarrow 44=4x\\\Rightarrow x=11

Money raised by each team = 94 + 11 = <em>$105</em>

Number of cars washed = <em>11</em>

4 0
3 years ago
Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a
8_murik_8 [283]

Answer:

  • a) 3/5·((-2)^n + 4·3^n)
  • b) 3·2^n - 5^n
  • c) 3·2^n + 4^n
  • d) 4 - 3 n
  • e) 2 + 3·(-1)^n
  • f) (-3)^n·(3 - 2n)
  • g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form a[n]=r^n, then it will satisfy ...

  r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}

Rearranging and dividing by r^{n-2}, we get the quadratic ...

  r^2-c_1r-c_2=0

The quadratic formula tells us values of r that satisfy this are ...

  r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

  a[n]=pr_1^n+qr_2^n

We can find p and q by solving the initial condition equations:

\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]

These have the solution ...

p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}

_____

Using these formulas on the first recurrence relation, we get ...

a)

c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n

__

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

  a[n]=(p+qn)r^n

The initial condition equations are now ...

\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]

and the solutions for p and q are ...

p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}

__

Using these formulas on problem (d), we get ...

d)

c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n

__

And for problem (f), we get ...

f)

c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0
3 years ago
In the figure shown, what is the measure of angle x? (5 points) Triangle ABC has measure of angle BAC equal to 50 degrees, and t
kiruha [24]

Answer:

Answer is 115 degree

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Leo's bank balances at the end of months 1, 2, and 3 are $1500, $1530, and $1560.60,
grandymaker [24]

Leo's balance after 9 months will be: $1757.49

Step-by-step explanation:

It is given that the balances follow a geometric sequence

First of all, we have to find the common ratio

Here

a_1 = 1500\\a_2 = 1530\\a_3 = 1560.60

Common ratio is:

r = \frac{a_2}{a_1} = \frac{1530}{1500} = 1.02\\r = \frac{a_3}{a_2} = \frac{1560.60}{1530} = 1.02

So r = 1.02

The general form for geometric sequence is:

a_n = a_1r^{n-1}

Putting the first term and r

a_n = 1500 . (1.02)^{n-1}

To find the 9th month's balance

Putting n=9

a_9 = 1500 . (1.02)^{9-1}\\= 1500.(1.02)^8\\=1757.4890

Rounding off to nearest hundredth

$1757.49

Hence,

Leo's balance after 9 months will be: $1757.49

Keywords: Geometric sequence, balance

Learn more about geometric sequence at:

  • brainly.com/question/10772025
  • brainly.com/question/10879401

#LearnwithBrainly

5 0
3 years ago
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