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Debora
13 days ago
9

A cylindrical cardboard tube with a diameter of 8 centimeters and a height of 20 centimeters is used to package a gift. A cylind

er has a height of 20 centimeters and a diameter of 8 centimeters. What is the approximate volume of the tube? Round to the nearest whole cubic centimeter. 1,005 cm3 1,340 cm3 3,351 cm3 4,021 cm3
Mathematics
2 answers:
tester [12.3K]13 days ago
6 0

Answer:

B

Detailed breakdown:

tester [12.3K]13 days ago
6 0

Response:

4021 cm^3

Detailed explanation:

The volume for a cylinder can be calculated using this formula.

V=\pi r^2h

where r represents the radius and h stands for the height.

In this case, the height of the cylinder is 20 centimeters, and its diameter measures 8 centimeters.

h= 20 cm

d= 8 cm

Now, we will input these values into the formula.

V=\pi (8 cm^2) * 20 cm

Firstly, compute the square.

8 cm^2= 8 cm * 8 cm= 64 cm

V=\pi * 64 cm^2*20 cm

Next, multiply 64 cm^2 with 20 cm

V=\pi *1280 cm^3

Then, calculate the product of 1280 cm^3 and pi

V=4021.2386 cm^3

Finally, round to the closest cubic centimeter. The digit 2 in the tenths position indicates to keep the number unchanged.

V= 4,021 cm^3

Thus, the volume is approximately 4,021 centimeters^3.

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Given matrix A below, and that A = B, find the value of the elements in B. A = 9 −2 3 2 17 0 3 22 8 b11 = b12 = b13 = b21 = b22
Svet_ta [12734]

Answer:

b_{11}=9,b_{12}=-2,b_{13}=3,b_{21}=2,b_{22}=17,b_{23}=0,b_{31}=3,b_{32}=22,b_{33}=8

Step-by-step explanation:

Review the provided matrix

A=\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]

Let matrix B be defined as

B=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]

It is stated that

A=B

\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]

By comparing the corresponding elements from both matrices, we derive

b_{11}=9,b_{12}=-2,b_{13}=3

b_{21}=2,b_{22}=17,b_{23}=0

b_{31}=3,b_{32}=22,b_{33}=8

Consequently, the needed values are b_{11}=9,b_{12}=-2,b_{13}=3,b_{21}=2,b_{22}=17,b_{23}=0,b_{31}=3,b_{32}=22,b_{33}=8.

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1 month ago
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Which equation shows the point-slope form of the line that passes through (3, 2) and has a slope of y plus StartFraction one-hal
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Answer:

B. y - 2 = 1/3(x - 3)

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1 month ago
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Mari has a part time job. She earns $7 an hour. She makes at most $143.50 a week. What is the greatest number of hours that she
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6 0
29 days ago
One of the industrial robots designed by a leading producer of servomechanisms has four major components. Components’ reliabilit
tester [12383]

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.

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1 month ago
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a) 278,382 oz b) 43 oz c) 64.1 oz
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