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3 years ago

9

On his first five math tests Colin receive the following scores: 82,90,93,62,and 81. What test score must : Burn on his six test

so that his average for all six of us 80?

1 answer:

ohaa [14]3 years ago

6 0

Answer:

The lowest he can get is a 72 to maintain an 80 as his average

Step-by-step explanation:

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Lydia bought 2 ½ gallons of paint and used 1 ½ gallons of paint. What fraction of the paint did she use?

grandymaker [24]

3/5 is used

2 1/2 gallons can be expressed as 5/2
1 1/2 gallons is the same as 3/2the fraction used is amount used over total amount

So (3/2)/(5/2) the 2 s cancel leaving you with 3/5

6 0

3 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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3 years ago

A system of two linear equations is graphed. The lines have the same slope, but different intercepts. How many solutions does th

rosijanka [135]

Hello,

lines are differents and parallel so 0 solutions
Answer A

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3 years ago

Why does putting a lid over red cabbage help it keeps it color

Dahasolnce [82]

Answer:

It is blocking out some bacteria that feast on the cabbage.making it lose it's color.

Step-by-step explanation:

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3 years ago

Leah has two same size rectangles divided into the same number of equal parts. One rectangle has 1/3 of the parts shaded, and th

steposvetlana [31]

<span>The number of parts is divisible by 3 and 5, so it must be a multiple of 15. So 15x1 = 15 is the smallest</span>

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3 years ago