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erik [133]
3 years ago
14

For what length of time does the log boat travel up the ride's conveyor belt?

Mathematics
2 answers:
Alenkasestr [34]3 years ago
7 0
<span>The <u>correct answer</u> is:

B) 30 seconds.

Explanation<span>:

The part of the graph where y increases represents the log boat traveling up the conveyor belt.

This starts at x=0 (0 seconds, when the ride begins) and ends at x=0.5 (0.5 minutes, or 30 seconds).  This means it takes 30-0 = 30 seconds.</span></span>
Goryan [66]3 years ago
4 0

the complete question in the attached figure

the solution is at the point  (0.5,70)

(0.5,70)--------------- > 0.50 minutes and 70 feet

0.50 minutes-------------- > 30 seconds

the answer is 30 seconds

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25/5 = x/2

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What is his average speed for end the entire trip 280 miles<br> ?
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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
3 years ago
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