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jolli1 [7]
3 years ago
13

6) (after 2.3) Consider the infinite system of linear equations in two variables given by ax + by = 0 where (a, b) moves along t

he unit circle in the plane. (a) How many solutions does this system have? (b) What is the smallest number of equations in the above system that have the same solution set? Write down two separate such linear systems, in vector form. (c) What happens to the infinite linear system if you add the equation 0x + 0y = 0 to it? (d) What happens to the infinite linear system if by accident one of the equations was recorded as ax + by = 0.00001? Explain all your answers in words.
Mathematics
1 answer:
Paraphin [41]3 years ago
6 0

Answer:

Step-by-step explanation:

Given infinite system of linear equations is ax + by = 0

when (a,b) moves along unit circle in plane.

a) system having unique system (0, 0)

Since two of equation in thus system will be

1.x+0.y=0\\x=0

and

0.x+1.y=0\\y=0

It is clear that x = 0, y= 0 is the only solution

b) Linear independent solution in this system gives some set of solutions

1.x+0.y=0\\\x=0

and

0.x+1.y=0\\y=0

Vector form is

\left[\begin{array}{ccc}1&0\\0&1\end{array}\right] =I

c) for this equation if add 0x +0y = 0 to system , Nothing will change

Because [0,0] satisfies that equation

d) If one of the equation is ax + by = 0.00001

where 0.00001 is small positive number

so, the system will be inconsistent

Therefore, the system will have no solution.

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Travelers who fail to cancel their hotel reservations when they have no intention of showing up are commonly referred to as no-s
notsponge [240]

Answer:

a) 0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) 0 is the most likely value for X.

Step-by-step explanation:

For each traveler who made a reservation, there are only two possible outcomes. Either they show up, or they do not. The probability of a traveler showing up is independent of other travelers. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

No-show rate of 10%.

This means that p = 0.1

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This means that n = 4

a) What is the probability that at least two of the four selected will turn to be no-shows?

This is P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{4,2}.(0.1)^{2}.(0.9)^{2} = 0.0486

P(X = 3) = C_{4,3}.(0.1)^{3}.(0.9)^{1} = 0.0036

P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001

P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0486 + 0.0036 + 0.0001 = 0.0523

0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) What is the most likely value for X?

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{4,0}.(0.1)^{0}.(0.9)^{4} = 0.6561

P(X = 1) = C_{4,1}.(0.1)^{1}.(0.9)^{3} = 0.2916

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P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001

X = 0 has the highest probability, which means that 0 is the most likely value for X.

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Step-by-step explanation:

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