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mamaluj [8]
2 years ago
12

"Use the given graph to determine the limit, if it exists.

Mathematics
2 answers:
meriva2 years ago
8 0
 \lim_{x \to 2} f(x)=3 ( x approaches 2 from the left )
\lim_{x \to 2} f ( x ) = -3 ( x approaches 2 from the right )

Dennis_Churaev [7]2 years ago
7 0

The <u>correct answer</u> is:


The limit of f(x) as <u>x approaches 2 from the left</u> is 3. This is represented algebraically by

\lim_{x \to 2} f(x) =3

The limit of f(x) as <u>x approaches 2 from the right</u> is -3. This is represented algebraically by

\lim_{x \to 2} f(x)=-3


Explanation:


From the left, we have a horizontal line approaching the value x=2. When we get to this point, the horizontal line stops at (2, 3). This means the limit is 3.


From the right, we have a horizontal line approaching the value x=2. When we get to this point, the horizontal line stops at (2, -3). This means the limit is -3.

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A college student is taking two courses. The probability she passes the first course is 0.73. The probability she passes the sec
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b) No, it's not independent.

c) 0.02

d) 0.59

e) 0.57

f) 0.5616

Step-by-step explanation:

To answer this problem, a Venn diagram should be useful. The diagram with the information of Event 1 and Event 2 is shown below (I already added the information for the intersection but we're going to see how to get that information in the b) part of the problem)

Let's call A the event that she passes the first course, then P(A)=.73

Let's call B the event that she passes the second course, then P(B)=.66

Then P(A∪B) is the probability that she passes the first or the second course (at least one of them) is the given probability. P(A∪B)=.98

b) Is the event she passes one course independent of the event that she passes the other course?

Two events are independent when P(A∩B) = P(A) * P(B)

So far, we don't know P(A∩B), but we do know that for all events, the next formula is true:

P(A∪B) = P(A) + P(B) - P(A∩B)

We are going to solve for P (A∩B)

.98 = .73 + .66 - P(A∩B)

P(A∩B) =.73 + .66 - .98

P(A∩B) = .41

Now we will see if the formula for independent events is true

P(A∩B) = P(A) x P(B)

.41 = .73 x .66

.41 ≠.4818

Therefore, these two events are not independent.

c) The probability she does not pass either course, is 1 - the probability that she passes either one of the courses (P(A∪B) = .98)

1 - P(A∪B) = 1 - .98 = .02

d) The probability she doesn't pass both courses is 1 - the probability that she passes both of the courses P(A∩B)

1 - P(A∩B) = 1 -.41 = .59

e) The probability she passes exactly one course would be the probability that she passes either course minus the probability that she passes both courses.

P(A∪B) - P(A∩B) = .98 - .41 = .57

f) Given that she passes the first course, the probability she passes the second would be a conditional probability P(B|A)

P(B|A) = P(A∩B) / P(A)

P(B|A) = .41 / .73 = .5616

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