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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?

For c(a.) Now we have:

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%.

For c(b.) Now we have:

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?

For d(a.) Now we have:

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%.

For d(b.) Now we have:

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%.

Final answer:

The probabilities of an auto arriving within a given time frame can be determined using the exponential distribution formula. When the rate of arrival is 50 per minute, the probability of an auto arriving within 6 seconds is approximately 0.9933 and within 3 seconds is approximately 0.9820. These probabilities increase with a higher rate of arrival and decrease with a lower rate of arrival.

Explanation:

To determine the probabilities of an auto arriving within a given time frame, we can use the exponential distribution formula. The exponential distribution is used to model the time until the next event occurs in a Poisson process, which is applicable in this scenario. The formula for the exponential distribution is: P(X <= t) = 1 - e-λt, where λ is the rate of arrival.

1. For a rate of arrival of 50 per minute, we can calculate the probability that the next auto will arrive within 6 seconds (0.1 minute) as: P(X <= 0.1) = 1 - e-50*0.1 = 1 - e-5 ≈ 0.9933.
2. Similarly, for the next auto to arrive within 3 seconds (0.05 minute), we can use the formula: P(X <= 0.05) = 1 - e-50*0.05 = 1 - e-2.5 ≈ 0.9820.
3. When the rate of arrival is 60 per minute, the probabilities will be slightly higher. The probability of the next auto arriving within 6 seconds would be approximately 0.9955, and within 3 seconds would be approximately 0.9838.
4. When the rate of arrival is 30 per minute, the probabilities will be slightly lower. The probability of the next auto arriving within 6 seconds would be approximately 0.9866, and within 3 seconds would be approximately 0.9641.

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Explanation:

Lateral Area

The lateral area is the area of the sides of the prism. If the faces are perpendicular to the bases, then each face is a rectangle. The area of each rectangle is the product of its length and width, generally the product of the height of the prism and the length of one edge of the base.

The total lateral area will then be the product of the height of the prism and the perimeter of the base.

Total Area

The total area is the sum of the lateral area (computed as above) and the area of the two bases of the prism. The formula for that area depends on the shape of the prism. (You have already seen formulas for the areas of triangles, rectangles, and other plane shapes. If not, they are readily available in your text or using a web search.)

The trade in price of her original car was $11,128.57.

Prices

Given that Suzanne has purchased a car with a list price of $23,860, and she traded in her previous car, which was a Dodge in good condition, and financed the rest of the cost for five years at a rate of 11.62%, compounded monthly, and the dealer gave her 85% of the listed trade-in price for her car, and she was also responsible for 8.11% sales tax on the difference between the list price and trade in price, a $1,695 vehicle registration fee, and a $228 documentation fee, if Suzanne makes a monthly payment of $455.96, to determine what was the trade in price of her original car, the following calculation must be made:

• 455.96 x 12 x 5 = 27,357.60
• X x (1 + 0.1162/12)^(12x5) = 27,357.60
• X x 1.009683^60 = 27,357.60
• 1.7828X = 27,357.60
• X = 27,357.60 / 1.7828
• X = 15,345
• 23,860 - 15,345 + (0.0811 x (23,860 - 15,345)) + 1695 + 228 = X
• 23,860 - 15,345 + 690.57 + 1695 + 228 = X
• 11,128.57 = X

Therefore, the trade in price of her original car was $11,128.57.

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Answer:

7 x 2/3 = 14/3 = 4 2/3

To find the area of a parallelogram, simply multiply the base times its height.

The value of the 23² number will be 529.

What is an expression?

Expression in maths is defined as the relation of numbers variables and functions by using mathematical signs like addition, subtraction, multiplication, and division.

In mathematics, expression is defined as the relationship of numbers, variables, and functions using mathematical signs such as addition, subtraction, multiplication, and division.

Given that there is a number 23 and is squared. The value of the 23² will be calculated as:-

E = 23²

E = 23 x 23

E = 529

Therefore, the value of the 23² number will be equal to 529.

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