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Westkost [7]
3 years ago
14

Suppose an assembly requires five components from five different vendors. To guarantee starting the assembly on time with 90 per

cent confidence, what must the service level be for each of the five components? (Assume the same service level for each component).
Engineering
1 answer:
spayn [35]3 years ago
6 0

Answer:

The service level for each component must be 97.91%

Explanation:

If we want a 90% confidence of starting on time, that means we need

P_{\mbox{starting on time}}=P_{\mbox{every component being ready on time}}=0.9\\

As the probability of each component being ready is independent from the others, that means that the probability of the 5 components being ready is equal to multiply each probability:

0.9=P_{\mbox{component 1 ready on time} } * P_{\mbox{component 2 ready on time} } *\\ P_{\mbox{component 3 ready on time} } * P_{\mbox{component 4 ready on time} } *\\P_{\mbox{component 5 ready on time} }

The probability of being ready on time is equal to the service level (in fraction), and all 5 are equal so we can say:

0.9=(\mbox{service level(in fraction)})^5\\\\\sqrt[5]{0.9} =\mbox{service level(in fraction)}=0.9791\\\mbox{In percentage}: \mbox{service level (in fraction)}*100 = 97.91\%

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C++ - Green Crud Fibonacci programThe following program is to be written with a loop. You are to write this program three times
Fynjy0 [20]

Answer:

Below is the required code:

Explanation:

Using for loop

#include <iostream>

using namespace std;

int main()

{

    //Initial crud size

    int init = 0;

    int newCrud;

    int next=0;

    //Number of days to simulate

    int no_days;

    int day;

    cout << "Enter initial amount of green crud: ";

    cin >> newCrud;

    cout << "Enter number of days to simulate: ";

    cin >> no_days;

    for (day = 10; day<=no_days; day++)

    {

         if (day % 10 == 0)

         {

             next = newCrud + init;

         }

             newCrud = init;

             init = next;

    }

    if (no_days < 5)

    cout << "\nCrud reproduce only after 5 days minimum.Hence the current amount is "

    << newCrud << " pounds.";

    else

    cout << "On day " << no_days << " you have " << init

    << " pounds of green crud." << endl;

    cout << "\nWould you like to continue? (y or n): ";

    cin >> ans;

         return 0;

}

Output:

         Enter initial amount of green crud: 5

         Enter number of days to simulate: 220

    On day 220 you have 10485760 pounds of green crud.

Using while loop

Program:

#include <iostream>

using namespace std;

int main()

{

    char ans='y';

    while (ans == 'Y' || ans == 'y')

    {

         //Initial crud size

         int init = 0;

         int newCrud;

         int next=0;

         //Number of days to simulate

         int no_days;

         int day;

         cout << "Enter initial amount of green crud:

         ";

         cin >> newCrud;

         cout << "Enter number of days to simulate:

         ";

         cin >> no_days;

         for (day = 10; day<=no_days; day++)

         {

             if (day % 10 == 0)

             {

                  next = newCrud + init;

             }

                  newCrud = init;

                  init = next;

         }

         if (no_days < 5)

         cout << "\nCrud reproduce only after 5 days

         minimum.Hence the current amount is "

         << newCrud << " pounds.";

         else

         cout << "On day " << no_days << " you have "

         << init

         << " pounds of green crud." << endl;

         cout << "\nWould you like to continue? (y or

         n): ";

         cin >> ans;

    }

    return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

Using do while loop

Program:

#include <iostream>

using namespace std;

int main()

{

    char ans;

    do

    {

         //Initial crud size

         int init = 0;

         int newCrud;

         int next=0;

         //Number of days to simulate

         int no_days;

         int day;

         cout << "Enter initial amount of green crud: ";

         cin >> newCrud;

         cout << "Enter number of days to simulate: ";

         cin >> no_days;

         for (day = 10; day<=no_days; day++)

         {

             if (day % 10 == 0)

             {

                  next = newCrud + init;

             }

                  newCrud = init;

                  init = next;

         }

         if (no_days < 5)

         cout << "\nCrud reproduce only after 5 days

         minimum.Hence the current amount is "

         << newCrud << " pounds.";

         else

         

         cout << "On day " << no_days << " you have " <<

         init << " pounds of green crud." << endl;

         cout << "\nWould you like to continue? (y or n):

         ";

         cin >> ans;

    } while (ans == 'Y' || ans == 'y');

    return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

7 0
3 years ago
A continuous and aligned fiber-reinforced composite having a cross-sectional area of 1130 mm2is subjected to an external tensile
lakkis [162]

Answer:

(a)  The force sustained by the matrix phase is 1802.35 N

(b) The modulus of elasticity of the composite material in the longitudinal direction Ed is 53.7 GPa

(c) The moduli of elasticity for the fiber and matrix phases is 124.8 GPa and 2.2 GPa respectively

Explanation:

Find attachment for explanation

8 0
3 years ago
Many BLANK apply trial and error to develop a product. Please Help! I have one hour to finish. 30 points
Julli [10]

Answer:

Entrepreneurs?

Explanation:

8 0
3 years ago
What does a peak flow meter allow you to assess?
Alex Ar [27]

Answer:

  peak flow and any engineering considerations related thereto

Explanation:

It should be no surprise that a peak flow meter will report peak flow, sometimes with important maximum-value, time-constant, or bandwidth limitations. There are many engineering issues related to flow rates. A peak flow meter can allow you to assess those issues with respect to the flows actually encountered.

Peak flow can allow you to assess adequacy of flow and whether there may be blockages or impediments to flow that reduce peak levels below expected values. An appropriate peak flow meter can help you assess the length of time that peak flow can be maintained, and whether that delivers sufficient volume.

It can also allow you to assess whether appropriate accommodation is made for unexpectedly high flow rates. (Are buffers or overflow tanks of sufficient size? Is there adequate protection against possible erosion? Is there adequate support where flow changes direction?)

3 0
3 years ago
The lattice constant of a simple cubic lattice is a0.
Oksi-84 [34.3K]

Answer:

A)The sketches for the required planes were drawn in the first attachment.

B)The sketches for the required directions were drawn in the second attachment.

To draw a plane in a simple cubic lattice, you have to follow these instructions:

1- the cube has 3 main directions called "a", "b" and "c" (as shown in the first attachment)

2- The coordinates of that plane are written as: π:(1/a₀ 1/b₀ 1/c₀) (if one of the coordinates is 0, for example (1 1 0), c₀ is ∞, therefore that plane never cross the direction c).

3- Identify the points a₀, b₀, and c₀ at the plane that crosses this main directions and point them in the cubic cell.

4- Join the points.

To draw a direction in a simple cubic lattice, you have to follow these instructions:

1- Identify the points a₀, b₀, and c₀ in the cubic cell.

2- Draw the direction as a vector-like (a₀ b₀ c₀).

7 0
2 years ago
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