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2 years ago

What is 4^2 - (3.1 + 6.4) + 4.5 pls show ur work

Mathematics

2 answers:

Ainat [17]2 years ago

7 0

Answer: 11
How:
4^2 - (3.1 + 6.4) + 4.5
16 - 9.5 + 4.5
6.5 + 4.5
= 11

melamori03 [73]2 years ago

5 0

4^ 2-(3.1+6.4)+4.5

16 − ( 3.1 + 6.4 ) + 4.5

16 − 9.5 + 4.5

6.5 + 4.5

= 11

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Answer:

mean = 7.8, median =5, and mode(s) =5

the shape of the distribution is positive skewed

Step-by-step explanation:

Find the mean, median and mode(s) for

2, 17, 5, 3, 28, 7, 5, 8, 5, 6, 2, 12, 10, 4, 3

the mean = sum of (2, 17, 5, 3, 28, 7, 5, 8, 5, 6, 2, 12, 10, 4, 3)/15 =

= 7.8

the median = 2, 2, 3, 3, 4, 5, 5, 5, 6, 7, 8, 10, 12, 17, 28, = the middle value, the 8th value falls in the middle = 5

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A right skewed histogram is a positive skewed histogram (which means the tail is longer the positive direction) that have its tail to the right.For a right skewed histogram the mean has a larger value that the median because the peak of the skewness fall in the median range then the mean follow to the right. and the data drive the mean value to the right which increase in value.

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3 years ago

How do you solve x2-8x+15=0

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3 years ago

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3 years ago

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

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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

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

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

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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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3/8

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