The cost to produce a product is modeled by the function f(x) = 3x2 − 6x + 11 where x is the number of products produced. complete the square to determine the minimum cost of producing this product.

Because this is a positive parabola that opens upwards, it natrually has a minimum value for the vertex. To complete the square we will first set the quadratic equal to 0 then move the constant over by subtraction. That gives us . One very important rule to follow about completing the square is that the leading coefficient ALWAYS be a +1. Our is a 3. So we will factor it out like this: . Next step is to take half the linear term, square it, and add it to both sides. Our linear term is 2. Half of 2 is 1, and 1 squared is 1. So we add 1 to both sides. BUT we cannot forget about that 3 we factored out. It refuses to be ignored. It is a multiplier. So when we add 1 into the set of parenthesis, we are REALLY adding in 3*1 = 3. We add 3 to the right side. All that looks like this: . The right side simplifies to -8, easy enough. What we did this for, this whole entire process, was to create a perfect square binomial on the left that will serve as the x coordinate of the vertex (actually the "h" since the vertex coordinates are (h, k)). Setting up that binomial looks like this now: . We move the -8 back over by addition, and find that the coordinates for the vertex are (1, 8). If x (h) is 1, that is "1 product" since x is products. If y (k) is 8, that is $8. What our vertex tells us in this particular situation is that it costs $8 to make 1 item.

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.