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Kobotan [32]
3 years ago
7

Please can someone help ill give the person who answers brainliest :>

Mathematics
1 answer:
Maru [420]3 years ago
5 0

Answer:

Q: i | P: ii | R: iii

Step-by-step explanation:

And I am gonna write a comment for you, if you just take my answer and get this right it means you have not understood the basics of graphing, and if you do not understand the simplest things you will struggle much more as you further your education.

<u>Keep studying and striving to become better educated!</u>

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Read 2 more answers
Autos arrive at a toll plaza located at the entrance to a bridge at a rate of 50 per minute during the​ 5:00-to-6:00 P.M. hour.
inna [77]

Answer:

a. The probability that the next auto will arrive within 6 seconds (0.1 minute) is 99.33%.

b. The probability that the next auto will arrive within 3 seconds (0.05 minute) is 91.79%.

c. What are the answers to (a) and (b) if the rate of arrival of autos is 60 per minute?

For c(a.), the probability that the next auto will arrive within 6 seconds (0.1 minute) is now 99.75%.

For c(b.), the probability that the next auto will arrive within 3 seconds (0.05 minute) is now 99.75%.

d. What are the answers to (a) and (b) if the rate of arrival of autos is 30 per minute?

For d(a.), the probability that the next auto will arrive within 6 seconds (0.1 minute) is now 95.02%.

For d(b.), the probability that the next auto will arrive within 3 seconds (0.05 minute) is now 77.67%.

Step-by-step explanation:

a. What is the probability that the next auto will arrive within 6 seconds (0.1 minute)?

Assume that x represents the exponential distribution with parameter v = 50,

Given this, we can therefore estimate the probability that the next auto will arrive within 6 seconds (0.1 minute) as follows:

P(x < x) = 1 – e^-(vx)

Where;

v = parameter = rate of autos that arrive per minute = 50

x = Number of minutes of arrival = 0.1 minutes

Therefore, we specifically define the probability and solve as follows:

P(x ≤ 0.1) = 1 – e^-(50 * 0.10)

P(x ≤ 0.1) = 1 – e^-5

P(x ≤ 0.1) = 1 – 0.00673794699908547

P(x ≤ 0.1) = 0.9933, or 99.33%

Therefore, the probability that the next auto will arrive within 6 seconds (0.1 minute) is 99.33%.

b. What is the probability that the next auto will arrive within 3 seconds (0.05 minute)?

Following the same process in part a, x is now equal to 0.05 and the specific probability to solve is as follows:

P(x ≤ 0.05) = 1 – e^-(50 * 0.05)

P(x ≤ 0.05) = 1 – e^-2.50

P(x ≤ 0.05) = 1 – 0.0820849986238988

P(x ≤ 0.05) = 0.9179, or 91.79%

Therefore, the probability that the next auto will arrive within 3 seconds (0.05 minute) is 91.79%.

c. What are the answers to (a) and (b) if the rate of arrival of autos is 60 per minute?

<u>For c(a.) Now we have:</u>

v = parameter = rate of autos that arrive per minute = 60

x = Number of minutes of arrival = 0.1 minutes

Therefore, we specifically define the probability and solve as follows:

P(x ≤ 0.1) = 1 – e^-(60 * 0.10)

P(x ≤ 0.1) = 1 – e^-6

P(x ≤ 0.1) = 1 – 0.00247875217666636

P(x ≤ 0.1) = 0.9975, or 99.75%

Therefore, the probability that the next auto will arrive within 6 seconds (0.1 minute) is now 99.75%.

<u>For c(b.) Now we have:</u>

v = parameter = rate of autos that arrive per minute = 60

x = Number of minutes of arrival = 0.05 minutes

Therefore, we specifically define the probability and solve as follows:

P(x ≤ 0.05) = 1 – e^-(60 * 0.05)

P(x ≤ 0.05) = 1 – e^-3

P(x ≤ 0.05) = 1 – 0.0497870683678639

P(x ≤ 0.05) = 0.950212931632136, or 95.02%

Therefore, the probability that the next auto will arrive within 3 seconds (0.05 minute) is now 99.75%.

d. What are the answers to (a) and (b) if the rate of arrival of autos is 30 per minute?

<u>For d(a.) Now we have:</u>

v = parameter = rate of autos that arrive per minute = 30

x = Number of minutes of arrival = 0.1 minutes

Therefore, we specifically define the probability and solve as follows:

P(x ≤ 0.1) = 1 – e^-(30 * 0.10)

P(x ≤ 0.1) = 1 – e^-3

P(x ≤ 0.1) = 1 – 0.0497870683678639

P(x ≤ 0.1) = 0.950212931632136, or 95.02%

Therefore, the probability that the next auto will arrive within 6 seconds (0.1 minute) is now 95.02%.

<u>For d(b.) Now we have:</u>

v = parameter = rate of autos that arrive per minute = 30

x = Number of minutes of arrival = 0.05 minutes

Therefore, we specifically define the probability and solve as follows:

P(x ≤ 0.05) = 1 – e^-(30 * 0.05)

P(x ≤ 0.05) = 1 – e^-1.50

P(x ≤ 0.05) = 1 – 0.22313016014843

P(x ≤ 0.05) = 0.7767, or 77.67%

Therefore, the probability that the next auto will arrive within 3 seconds (0.05 minute) is now 77.67%.

8 0
3 years ago
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