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mixer [17]
2 years ago
5

g The completion times for a job task range from 11.8 minutes to 19.4 minutes and are thought to be uniformly distributed. What

is the probability that it will require between 12.8 and 16.8 minutes to perform the task
Mathematics
1 answer:
nevsk [136]2 years ago
5 0

Answer: 0.5263

Step-by-step explanation:

Given : The completion times for a job task range from 11.8 minutes to 19.4 minutes .

The density function for uniform distribution function for interval [a,b] :-

f(x)=\dfrac{1}{b-a}

Then the density function for the given situation:-

f(x)=\dfrac{1}{19.4-11.8}=\dfrac{1}{7.6}

The required interval : 16.8-12.8=4

Now, the probability that it will require between 12.8 and 16.8 minutes to perform the task will be :-

\dfrac{4}{7.6}=0.526315789474\approx0.5263

You might be interested in
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:

<u>Component Reliabilities:</u>

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:

<u>Reliability of the Robot with backup of component 1</u> 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

<u>Similarly, to find out the reliability of component 2:</u>

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

<u>Reliability of the Robot with backup of component 3 can be computed as:</u>

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

<u>Reliability of the Robot with backup of component 4 can be computed as:</u>

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 <u>highest reliability can be achieved by adding a backup of component 4 with a value of 0.8664</u>. So, <u>Component 4 should get the backup in order to achieve the highest reliability.</u>

<u></u>

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 <u>chances of failure</u> 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 <u>reliability for each of the component & its backup</u> 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

<u>The reliability of the robot with backups</u> 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

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

4 0
2 years ago
Which of the following functions shown in the table below is not an exponential function?
kramer

Answer:

B. f(x)

D. h(x)

Step-by-step explanation:

An exponential function will have a common ratio.

We check and see if the consecutive terms has a common ratio.

For f(x),

\frac{5}{2}  \ne \frac{10}{5}

This is not an exponential sequence.

For g(x), we have:

\frac{2}{1}  =  \frac{4}{2}  =  \frac{8}{4}  = 2

For h(x),

\frac{1.25}{1}  \ne \frac{1.5}{1.25}

This is not an exponential sequence

For k(x),

\frac{16}{64}  =  \frac{4}{16}  =  \frac{1}{4}  =  \frac{0.25}{1}

6 0
2 years ago
489 to the nearest ten​
sammy [17]
8 is in the tenth position.

In order to round up, the number behind it (in this case. 9) must be one of the numbers of 5-9. Because 9 meets this requirement, you can round 8 to 9 and this will make 9 to 0.

Your answer is 490
4 0
2 years ago
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HURRY PLEASE When 5 and 6 are multiplied by the same factor, how do the ratios compare to the ratio 5:6? The ratios are equivale
erma4kov [3.2K]

Answer:

The ratios are equivalent.

Step-by-step explanation:

If you multiply both 5 and 6, the ratio stays the same.

5:6 = 10:12 = 15:18 = 20:24 etc.

You can try this by dividing the larger number by the smaller number. You'll always get 1.2 as the "relationship", or ratio, is preserved.

24/20 = 1.2

15/18 = 1.2

10/12 = 1.2

6/5 = 1.2

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Which are true statements about a translation? Select all that apply. The image is congruent to the pre-image. The image could b
Novay_Z [31]

Answer:

a} The image is congruent to the pre-image.

c} The image could be moved left or right.

d} The image could be moved up or down.

Step-by-step explanation:

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