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LenKa [72]
3 years ago
5

Which statement is true about the discontinuities of the function f(x)?

Mathematics
2 answers:
DedPeter [7]3 years ago
8 0

Answer:

f(x) = (x + 1)/ (6x^2 - 7x - 3)

= (x + 1)( / (6x^2 + 2x - 9x  - 3)

= (x + 1) / (2x(3x + 1) - 3(3x + 1))

= (x + 1) / (2x - 3)(3x + 1)

Now x = 3/2 and x = -1/3 both make te denominator zero so these   are both asymptotes.

Read more on Brainly.com - brainly.com/question/15296078#readmore

Step-by-step explanation:

Tema [17]3 years ago
6 0

Answer:

There are asymptotes at x = three-halves and x = negative one-third.

Step-by-step explanation:

f(x) = (x + 1)/ (6x^2 - 7x - 3)

= (x + 1)( / (6x^2 + 2x - 9x  - 3)

= (x + 1) / (2x(3x + 1) - 3(3x + 1))

= (x + 1) / (2x - 3)(3x + 1)

Now x = 3/2 and x = -1/3 both make te denominator zero so these   are both asymptotes.

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Motor vehicles sold to individuals are classified as either cars or light trucks (including SUVs) and as either domestic or impo
Rudik [331]

Answer:

a) Probability that vehicle is Light Truck is 0.69

b) Probability that vehicle is imported car is 0.08 or 8%.

Step-by-step explanation:

Part a)

Event that vehicle is a car is given by A. Since, the vehicles can only be classified as cars or light trucks, the event that a vehicle will be light truck will be compliment of A.

i.e. Event that vehicle is a Light Truck = A^{c}

It is given that in recent year 69% of vehicles sold were light trucks, 78% were domestic, and 55% were domestic light trucks.

So, from here we can say that, if a vehicle is randomly selected from all the vehicles the probability that it would be a light truck will be:

P(Vehicle will be Light Truck) = P(A^{c}) = 69% = 0.69

Part b)

Event that vehicle is imported is given by B. We need to find the probability the a randomly chosen vehicle is an imported car i.e. we have to find probability of occurrence of events A and B together, which will be denoted as: P(A and B)

Since, P(A^{c}) = 69% = 0.69

P(A) = 1 - P(A^{c}) = 1 - 0.69 = 0.31

It is also given that 78% of vehicles were domestic. This means, the percentage/probability of imported vehicles is:

P(B) = 1 - 0.78 = 0.22

55% of vehicles were domestic light trucks. This can be expressed as:

P(A^{c}\&B^{c}), the compliment of this event will give us P(A or B).

i.e.

P(A or B) = 1 - P(A^{c} \&B^{c}) = 1 - 0.55 = 0.45

According to the addition rule of probabilities:

P(A or B) = P(A) + P(B) - P(A and B)

Substituting the calculated values gives us:

0.45 = 0.31 + 0.22 - P(A and B)

P(A and B) = 0.08

This means, the probability that vehicle is imported car is 0.08 or 8%.

4 0
3 years ago
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