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2 years ago

Amar cycles at a speed of 18km/h.

Mathematics

1 answer:

damaskus [11]2 years ago

7 0

Answer:

The distance between the two villages is 16.5 Km

Step-by-step explanation:

Constant Speed Motion

It's a type of motion in which the distance of an object changes by an equal amount in every equal period of time.

If v is the constant speed, the object travels a distance x in a time t, given by the equation:

x=vt

Amar cycles at v=18 Km/h for t=55 minutes. We need to calculate the distance traveled between the two villages.

Since the speed and the time are given in different units, we convert the time to hours, recalling that 1 hour=60 minute.

t=55 min = 55/60 hours

For the sake of precision, we won't operate the division so far. Compute the distance:

x=18 *55/60=16.5 Km

The distance between the two villages is 16.5 Km

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The time between arrivals of small aircraft at a county airport is exponentially distributed with a mean of one hour. Round the

navik [9.2K]

Answer:

a) 0.019

b) 0.563

c) x = 1.966 hours

Step-by-step explanation:

E(X) = 1

Exponential random variable's probability function is given as

P(X=x) = λ e^(-λ.x)

The cumulative distribution function is given as

P(X ≤ x) = 1 - e^(-λ.x)

a) The time between the arrivals of small aircraft at a county airport that is exponentially distributed.

But the number of planes that land every hour will be obtained using the Poisson distribution formula.

It is the best for discrete systems.

Poisson distribution formula is given as

P(X = x) = (e^-λ)(λˣ)/x!

λ = 1 aircraft per hour.

The probability that more than three aircraft arrive within an hour = P(X > 3)

P(X > 3) = 1 - P(X ≤ 3) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]

P(X > 3) = 1 - 0.98101 = 0.01899 = 0.019 to 3 d.p

b) If 30 separate one-hour intervals are chosen, what Is the probability that no interval contains more than three arrivals

Probability of one 1-hour interval not containing more than 3 arrivals = 1 - P(X > 3)

= 1 - 0.01899 = 0.98101

Probability that thirty 1-hour intervals will not contain more than 3 arrivals = (0.98101)³⁰ = 0.5626 = 0.563 to 3 d.p

c) Determine the length of an interval of time (In hours) such that the probability that no arrivals occur during the interval is 0.14

We can now use the cumulative distribution function for exponential random variable for this

P(X ≤ x) = 1 - e^(-λ.x)

P(X > x) = 1 - P(X ≤ x)

P(X > x) = e^(-λ.x)

λ = 1, x = ?,

0.14 = e⁻ˣ

e⁻ˣ = 0.14

In e⁻ˣ = In 0.14 = -1.966

-x = -1.966

x = 1.966 hours

Hope this Helps!!!

4 0

2 years ago

A cell tower is located 3 miles east and 4 miles north of the center of a small town. The cell tower has a coverage radius of 3

butalik [34]

Answer:

The answer is below

Step-by-step explanation:

A cell tower is located 3 miles east and 4 miles north of the center of a small town. The cell tower has a coverage radius of 3 miles.

a) Start by drawing a diagram of this situation. Your diagram might include a coordinate plane, the cell tower, and a circle representing the cell tower's coverage boundary.

a) A house is located 5 miles north of the center of the town and is to the east of the cell tower. If the house lies on the boundary of the cell tower's coverage, how far east of the center of the town is the house?

Answer:

Let the center of the town represent the origin (0,0). Also 1 unit = 1 mile. Since the cell tower is located 3 miles east and 4 miles north of the center of a small town, it is represented by A(3, 4)

The cell tower has a coverage of radius 3 miles. This can be represented by a circle with equation:

(x - a)² + (y - b)² = r². where (a,b) is the center of the circle and r is the radius. Hence:

(x - 3)² + (y - 4)² = 3²

(x - 3)² + (y - 4)² = 9

The diagram is drawn using geogebra.

b) The house is 5 miles north. It can be represented by y = 5 line.

To find the distance east of the house we have to substitute y = 5 and solve for x, hence:

(x - 3)² + (y - 4)² = 9

(x - 3)² + (5 - 4)² = 9

(x - 3)² + 1 = 9

(x - 3)² = 8

(x - 3) = √8

x - 3 = ±2.83

x = 3 ± 2.83

x = 5.83 or 1.83

Since it is to the east to the cell tower, hence x = 5.83.

Therefore the house is located 5.83 miles to the east of the cell tower

8 0

2 years ago

A mill processes 27 tons of flour in 3 hours. How many tons of flour can the mill process in 8 hours if the rate of production s

Delvig [45]

There are a few ways you can go about solving this question. One way is to use the given information to find how many tons of flour can be processed in one hour. If 27 tons can be processed in 3 hours, we can do 27 divided by 3 to find that 9 tons of flour can be processed per hour. Then, if we want to see how many tons of flour can be processed in 8 hours, we can multiply 9 tons by 8 hours to get a total of 72 tons of flour.

I hope this helps.

5 0

2 years ago

A model that describes the population of a fishery in which harvesting takes place at a constant rate is given by dP dt = kP − h

Vadim26 [7]

It increases at T because the behavior increased its speed

8 0

2 years ago

One of the industrial robots designed by a leading producer of servomechanisms has four major components. Components’ reliabilit

Ivahew [28]

Answer:

a) Reliability of the Robot = 0.7876

b1) Component 1: 0.8034

Component 2: 0.8270

Component 3: 0.8349

Component 4: 0.8664

b2) Component 4 should get the backup in order to achieve the highest reliability.

c) Component 4 should get the backup with a reliability of 0.92, to obtain the highest overall reliability i.e. 0.8681.

Step-by-step explanation:

Component Reliabilities:

Component 1 (R1) : 0.98

Component 2 (R2) : 0.95

Component 3 (R3) : 0.94

Component 4 (R4) : 0.90

a) Reliability of the robot can be calculated by considering the reliabilities of all the components which are used to design the robot.

Reliability of the Robot = R1 x R2 x R3 x R4

= 0.98 x 0.95 x 0.94 x 0.90

Reliability of the Robot = 0.787626 ≅ 0.7876

b1) Since only one backup can be added at a time and the reliability of that backup component is the same as the original one, we will consider the backups of each of the components one by one:

Reliability of the Robot with backup of component 1 can be computed by first finding out the chance of failure of the component along with its backup:

Chance of failure = 1 - reliability of component 1

= 1 - 0.98

= 0.02

Chance of failure of component 1 along with its backup = 0.02 x 0.02 = 0.0004

So, the reliability of component 1 and its backup (R1B) = 1 - 0.0004 = 0.9996

Reliability of the Robot = R1B x R2 x R3 x R4

= 0.9996 x 0.95 x 0.94 x 0.90

Reliability of the Robot = 0.8034

Similarly, to find out the reliability of component 2:

Chance of failure of component 2 = 1 - 0.95 = 0.05

Chance of failure of component 2 and its backup = 0.05 x 0.05 = 0.0025

Reliability of component 2 and its backup (R2B) = 1 - 0.0025 = 0.9975

Reliability of the Robot = R1 x R2B x R3 x R4

= 0.98 x 0.9975 x 0.94 x 0.90

Reliability of the Robot = 0.8270

Reliability of the Robot with backup of component 3 can be computed as:

Chance of failure of component 3 = 1 - 0.94 = 0.06

Chance of failure of component 3 and its backup = 0.06 x 0.06 = 0.0036

Reliability of component 3 and its backup (R3B) = 1 - 0.0036 = 0.9964

Reliability of the Robot = R1 x R2 x R3B x R4

= 0.98 x 0.95 x 0.9964 x 0.90

Reliability of the Robot = 0.8349

Reliability of the Robot with backup of component 4 can be computed as:

Chance of failure of component 4 = 1 - 0.90 = 0.10

Chance of failure of component 4 and its backup = 0.10 x 0.10 = 0.01

Reliability of component 4 and its backup (R4B) = 1 - 0.01 = 0.99

Reliability of the Robot = R1 x R2 x R3 x R4B

= 0.98 x 0.95 x 0.94 x 0.99

Reliability of the Robot = 0.8664

b2) According to the calculated values, the highest reliability can be achieved by adding a backup of component 4 with a value of 0.8664. So, Component 4 should get the backup in order to achieve the highest reliability.

c) 0.92 reliability means the chance of failure = 1 - 0.92 = 0.08

We know the chances of failure of each of the individual components. The chances of failure of the components along with the backup can be computed as:

Component 1 = 0.02 x 0.08 = 0.0016

Component 2 = 0.05 x 0.08 = 0.0040

Component 3 = 0.06 x 0.08 = 0.0048

Component 4 = 0.10 x 0.08 = 0.0080

So, the reliability for each of the component & its backup is:

Component 1 (R1BB) = 1 - 0.0016 = 0.9984

Component 2 (R2BB) = 1 - 0.0040 = 0.9960

Component 3 (R3BB) = 1 - 0.0048 = 0.9952

Component 4 (R4BB) = 1 - 0.0080 = 0.9920

The reliability of the robot with backups for each of the components can be computed as:

Reliability with Component 1 Backup = R1BB x R2 x R3 x R4

= 0.9984 x 0.95 x 0.94 x 0.90

Reliability with Component 1 Backup = 0.8024

Reliability with Component 2 Backup = R1 x R2BB x R3 x R4

= 0.98 x 0.9960 x 0.94 x 0.90

Reliability with Component 2 Backup = 0.8258

Reliability with Component 3 Backup = R1 x R2 x R3BB x R4

= 0.98 x 0.95 x 0.9952 x 0.90

Reliability with Component 3 Backup = 0.8339

Reliability with Component 4 Backup = R1 x R2 x R3 x R4BB

= 0.98 x 0.95 x 0.94 x 0.9920

Reliability with Component 4 Backup = 0.8681

Component 4 should get the backup with a reliability of 0.92, to obtain the highest overall reliability i.e. 0.8681.

4 0

2 years ago