Find the volume of a right circular cone that has a height of 12.2 cm and a base with a circumference of 18.5 cm. Round your ans

Answer:

a) Robot Reliability = 0.7876

b1) Component 1: 0.8034

    Component 2: 0.8270

    Component 3: 0.8349

    Component 4: 0.8664

b2) To maximize overall reliability, Component 4 should be backed up.

c) To achieve the highest reliability of 0.8681, backup for Component 4 with a reliability of 0.92 should be implemented.

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) The reliability of the robot can be determined by calculating the reliabilities of the individual components that constitute the robot.

Robot Reliability = R1 x R2 x R3 x R4

                                      = 0.98 x 0.95 x 0.94 x 0.90

Robot Reliability = 0.787626 ≅ 0.7876

b1) As only a single backup can be used at once, and its reliability matches that of the original, we evaluate each component's backup sequentially:

Robot Reliability with Component 1 backup is calculated by first assessing the failure probability of the component plus its backup:

Failure probability = 1 - R1

                      = 1 - 0.98

                      = 0.02

Combined failure probability for Component 1 and backup = 0.02 x 0.02 = 0.0004

Thus, reliability of combined Component 1 and backup (R1B) = 1 - 0.0004 = 0.9996

Robot Reliability = R1B x R2 x R3 x R4

                                         = 0.9996 x 0.95 x 0.94 x 0.90

Robot Reliability = 0.8034

To determine reliability of Component 2:

Failure probability for Component 2 = 1 - 0.95 = 0.05

Combined failure probability of Component 2 and backup = 0.05 x 0.05 = 0.0025

Reliability of Component 2 with backup (R2B) = 1 - 0.0025 = 0.9975

Robot Reliability = R1 x R2B x R3 x R4

                = 0.98 x 0.9975 x 0.94 x 0.90

Robot Reliability = 0.8270

Robot Reliability with backup of Component 3 calculates as follows:

Failure probability for Component 3 = 1 - 0.94 = 0.06

Combined failure probability of Component 3 and backup = 0.06 x 0.06 = 0.0036

Reliability for Component 3 with backup (R3B) = 1 - 0.0036 = 0.9964

Robot Reliability = R1 x R2 x R3B x R4  

                = 0.98 x 0.95 x 0.9964 x 0.90

Robot Reliability = 0.8349

Robot Reliability with Component 4 backup calculates as:

Failure probability for Component 4 = 1 - 0.90 = 0.10

Combined failure probability of Component 4 and backup = 0.10 x 0.10 = 0.01

Reliability for Component 4 and backup (R4B) = 1 - 0.01 = 0.99

Robot Reliability = R1 x R2 x R3 x R4B

                                      = 0.98 x 0.95 x 0.94 x 0.99

Robot Reliability = 0.8664

b2) The best reliability is achieved with the backup of Component 4, yielding a value of 0.8664. Thus, Component 4 is the best candidate for backup to optimize reliability.

c) A reliability of 0.92 indicates a failure probability of = 1 - 0.92 = 0.08

We can compute the probability of failure for each component along with its backup:

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

Thus, the reliabilities for each component and its backup become:

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

Reliability of robot including backups for each of the components can be calculated as:

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

              = 0.9984 x 0.95 x 0.94 x 0.90

Reliability with Backup for Component 1 = 0.8024

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

              = 0.98 x 0.9960 x 0.94 x 0.90

Reliability with Backup for Component 2 = 0.8258

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

              = 0.98 x 0.95 x 0.9952 x 0.90

Reliability with Backup for Component 3 = 0.8339

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

              = 0.98 x 0.95 x 0.94 x 0.9920

Reliability with Backup for Component 4 = 0.8681

To maximize overall reliability, Component 4 should be backed up at a reliability of 0.92, achieving an overall reliability of 0.8681.

Answer: 0.31

Step-by-step explanation:

Let A signifies the event of the Tampa Bay Buccaneers scoring a touchdown on their initial drive while B signifies the occurrence of their defense achieving 3 or more sacks during the game.

Given: P(A)=0.14     P(B) = 0.31     P(A or B)=0.14

Formula: P(A and B)= P(A) + P(B) - P(A or B)

Now, we calculate the probability of both scoring a touchdown on the initial drive and achieving 3 or more sacks in the game as follows:-

P(A and B)= 0.14 + 0.31 - 0.14=0.31

Therefore, the required probability is: 0.31

Answer:

She is mistaken

Step-by-step explanation:

Step 1: input

135 for x in both equations

Step 2: Solve those equations. This will yield y = 5375 for the first equation and y = 4720 for the second, indicating that she is incorrect

The percentage rise in Boris's car insurance costs is 249%. Step-by-step explanation: To find the percentage increase in Boris's car insurance, we need to apply a mathematical formula. The formula is {(new amount - old amount) / old amount} * 100%. Given the previous data, the old amount is £256, and the new amount is £894. Hence, the percentage increase results in (894 - 256) / 256 * 100% = 638 / 256 * 100 = 249.21875, which rounds to 249%.

Answer:

a) 47.55

b) 58

c) 47.88

Step-by-step explanation:

Considering that the order sizes are uniformly distributed across the range

$25 ( a ) to $80 ( b )

a) Calculate the measurement for the initial order size, applying a random number of 0.41

The parameters for the normal distribution are a = 25, b = 80

The size or value of the order is calculated as follows:  = a + random number ( b - a )

                                = 25 + 0.41 ( 80 - 25 )

                                =  47.55

b) The last order's generated value using the random number (0.6)

= a + random number ( b - a )

= 25 + 0.6 ( 80 - 25 )

= 25 + 33 = 58

c) Average order size

= ∑ order 1 + order 2 + ----- + order 10  ) / 10

= (47.55 +...... + 58 ) / 10

= 478.8 / 10 = 47.88