F(-3)= 6x^4-3x^3-20x^2+30x+3

The weekly revenue for a company is r = 3p^2 + 60p + 1060, where p is the price of the company's product. what price will result in a revenue of $1200.

Problem situation: caren is making rice and beans. she can spend no more than $10 on ingredients. she buys one bag of rice for $4.99. beans cost $0.74 per pound. how many pounds of beans, x, can she buy? inequality that represents this situation: 10≥4.99+0.74x drag each number to show if it is a solution to both the inequality and the problem situation, to the inequality only, or if it is not a solution.

Each of the following data sets has a mean of x = 10. (i) 8 9 10 11 12 (ii) 7 9 10 11 13 (iii) 7 8 10 12 13 (a) without doing any computations, order the data sets according to increasing value of standard deviations. (i), (iii), (ii) (ii), (i), (iii) (iii), (i), (ii) (iii), (ii), (i) (i), (ii), (iii) (ii), (iii), (i) (b) why do you expect the difference in standard deviations between data sets (i) and (ii) to be greater than the difference in standard deviations between data sets (ii) and (iii)? hint: consider how much the data in the respective sets differ from the mean. the data change between data sets (i) and (ii) increased the squared difference îł(x - x)2 by more than data sets (ii) and (iii). the data change between data sets (ii) and (iii) increased the squared difference îł(x - x)2 by more than data sets (i) and (ii). the data change between data sets (i) and (ii) decreased the squared difference îł(x - x)2 by more than data sets (ii) and (iii). none of the above