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3 years ago

Please help with this limit :(((

Mathematics

2 answers:

salantis [7]3 years ago

7 0

Answer:

doesn't exist is the answer.

V125BC [204]3 years ago

6 0

Does not exist is the correct answer

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60% expressed as a ration is StartFraction 60 Over 100 EndFraction. What is this as a fraction in its reduced form?

Leno4ka [110]

Answer:

3/5

Step-by-step explanation:

60% means 60 out of 100. For example if you get 60 marks in an exam of 100 marks. This means you got 60% marks.

Which is 60/100. This fraction can further be reduced to 6/10 [divide both numerator and denominator by 10)

6/10 = 3/5 [divide both numerator and denominator by 2)

So, the reduced form of 60% =3/5

4 0

3 years ago

Suppose a certain computer virus can enter a system through an email or through a webpage. There is a 40% chance of receiving th

DedPeter [7]

Answer:

P = 0.42

Step-by-step explanation:

This probability problem can be solved by building a Venn like diagram for each probability.

I say that we have two sets:

-Set A, that is the probability of receiving this virus through the email.

-Set B, that is the probability of receiving it through the webpage.

The most important information in these kind of problems is the intersection. That is, that he virus enters the system simultaneously by both email and webpage with a probability of 0.17. It means that $A \cap B$ = 0.17.

By email only

The problem states that there is a 40 chance of receiving it through the email. It means that we have the following equation:

$A + (A \cap B) = 0.40$

$A + 0.17 = 0.40$

$A = 0.23$

where A is the probability that the system receives the virus just through the email.

The problem states that there is a 40% chance of receiving it through the email. 23% just through email and 17% by both the email and the webpage.

By webpage only

There is a 35% chance of receiving it through the webpage. With this information, we have the following equation:

$B + (A \cap B) = 0.35$

$B + 0.17 = 0.35$

$B = 0.18$

where B is the probability that the system receives the virus just through the webpage.

The problem states that there is a 35% chance of receiving it through the webpage. 18% just through the webpage and 17% by both the email and the webpage.

What is the probability that the virus does not enter the system at all?

So, we have the following probabilities.

- The virus does not enter the system: P

- The virus enters the system just by email: 23% = 0.23

- The virus enters the system just by webpage: 18% = 0.18

- The virus enters the system both by email and by the webpage: 17% = 0.17.

The sum of the probabilities is 100% = 1. So:

P + 0.23 + 0.18 + 0.17 = 1

P = 1 - 0.58

P = 0.42

There is a probability of 42% that the virus does not enter the system at all.

5 0