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3 years ago

lessils used 25% of half a gallon (64 ounces) of milk for a recipe how many ounces of milk where left

Mathematics

1 answer:

GalinKa [24]3 years ago

3 0

Answer:

There were 48 ounces of milk left

Step-by-step explanation:

∵ Lessils used 25% of half a gallon of milk

→ Assume that the total amount of the milk is 25%

∴ The amount of milk left = 100% - 25% = 75%

∵ Half gallon = 64 ounces

→ That means the amount of milk left is 75% of 64 ounces

∴ The amount of milk left = 75% × 64

→ Divide 75% by 100 to change it to decimal

∵ 75% = 75 ÷ 100 = 0.75

∴ The amount of milk left = 0.75 × 64

∴ The amount of milk left = 48 ounces

∴ There were 48 ounces of milk left

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2 years ago

Suppose a certain manufacturing company produces connecting rods for 4- and 6-cylinder automobile engines using the same product

gregori [183]

Answer:

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

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

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

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Step-by-step explanation:

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