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Yuri [45]
2 years ago
5

An inspector found that 4 out of 116 parts were defective. What percent of the parts were defective?

Mathematics
2 answers:
daser333 [38]2 years ago
4 0
A. 3.45%
The formula for a proportion is x/n, "n" is the total of objects and "x" is the number os successes. In this case "defective" is the success because you want to know the percentage for defective parts, so 4/116= 0.0345 and you have to multiply the result by 100 and you get your answer in percentage.
Margarita [4]2 years ago
3 0

Answer:

A. 3.45%

Step-by-step explanation:

Given,

Total parts = 116,

Defective parts = 4,

Thus, the percentage of the parts that were defective

=\frac{\text{D efective parts}}{\text{Total parts}}\times 100

=\frac{4}{116}\times 100

=\frac{400}{116}

=3.44827586207\%

\approx 3.45\%

Option A is correct.

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Suppose a certain manufacturing company produces connecting rods for 4- and 6-cylinder automobile engines using the same product
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Generally the constraint that sets next week are shown below

Generally the constrain that sets next week maximum production of connecting rod for 4 cylinder  to  W_4 or  0 is  

     x_4 \le W_4 *  s_4

    x_4 \le 5000 *  s_4

Generally the constrain that sets next week maximum production of connecting rod for 6 cylinder  to  W_6 or  0  is  

     x_6 \le W_6 *  s_6

     x_6 \le 8,000 *  s_6

Generally the constrain that limits the production of connecting rods  for both 4 cylinder and 6 cylinders  is

     x_4 \le W_4 *  s_6

=>   x_4 \le 5000 *  s_6

     x_4 \le W_6 *  s_4

=>    x_4 \le 8000 *  s_4

     s_4 + s_6 = 1

The minimum cost of production for next week is  

   U  =  M_4 *  x_4 + M_6 * x_6 + C_4 * s_4 + C_6 * s_6

=>  U  =  13x_4 + 16x_6 + 2000 s_4 + 3500 s_6

Step-by-step explanation:

The cost for the four cylinder production line is  C_4 =  \$2,100

The cost for the six cylinder production line is  C_6 = \$3,500

The manufacturing cost for each four cylinder is  M_4= \$13

 The manufacturing cost for each six cylinder is M_6= \$16

  The weekly production capacity for 4 cylinder connecting rod is W_4 = 5,000

   The weekly production capacity for 6 cylinder connecting rod is W_6 = 8,000

Generally the constraint that sets next week are shown below

Generally the constrain that sets next week maximum production of connecting rod for 4 cylinder  to  W_4 or  0 is  

     x_4 \le W_4 *  s_4

    x_4 \le 5000 *  s_4

Generally the constrain that sets next week maximum production of connecting rod for 6 cylinder  to  W_6 or  0  is  

     x_6 \le W_6 *  s_6

     x_6 \le 8,000 *  s_6

Generally the constrain that limits the production of connecting rods  for both 4 cylinder and 6 cylinders  is

     x_4 \le W_4 *  s_6

=>   x_4 \le 5000 *  s_6

     x_4 \le W_6 *  s_4

=>    x_4 \le 8000 *  s_4

     s_4 + s_6 = 1

The minimum cost of production for next week is  

   U  =  M_4 *  x_4 + M_6 * x_6 + C_4 * s_4 + C_6 * s_6

=>  U  =  13x_4 + 16x_6 + 2000 s_4 + 3500 s_6

3 0
3 years ago
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