answer:To solve the problem, we start by understanding that the average of a set of numbers is the sum of the numbers divided by the count of the numbers. Given that the average of 10 numbers is 85, we can express this relationship as: [ frac{S}{10} = 85 ] where S is the sum of the 10 numbers. From this equation, we can solve for S to find the total sum of these numbers: [ S = 85 times 10 = 850 ] This means the sum of all 10 numbers is 850. When the numbers 70 and 76 are removed, the sum of the remaining numbers is: [ S - 70 - 76 = 850 - 70 - 76 = 704 ] Now, with 8 numbers remaining (since 2 have been removed), the average of these numbers is the sum of these numbers divided by 8: [ frac{704}{8} = 88 ] Therefore, the average of the remaining 8 numbers is boxed{88}.

answer:**Solution**: To rewrite the proposition "The diagonals of a rhombus are perpendicular to each other" in the form of "If ..., then ...", we can express it as: If a quadrilateral is a rhombus, then the diagonals of this quadrilateral are perpendicular to each other. Therefore, the rewritten proposition is: boxed{text{If a quadrilateral is a rhombus, then the diagonals of this quadrilateral are perpendicular to each other.}}

answer:Solution: overrightarrow {a}cdot overrightarrow {b} = 3 × 2 × cos frac {2π}{3} = -3, ∴ overrightarrow {a}cdot (overrightarrow {a}-2overrightarrow {b}) = overrightarrow {a}^{2} - 2 overrightarrow {a}cdot overrightarrow {b} = 9 - 2 × (-3) = 15. Hence, the answer is: boxed{D}. First, calculate overrightarrow {a}cdot overrightarrow {b}, then use the properties of the dot product of planar vectors to compute the result. This question tests your understanding of the dot product of planar vectors and is considered a basic problem.

answer:[Analysis] This problem tests our ability to use derivatives to study the equation of a tangent line at a point on a curve. It requires mathematical transformation and thinking skills. The key to solving this problem is to convert it into a relationship between sets. This is a moderately difficult problem. [Solution] First, we find the derivatives of the given functions: f'(x)=-e^{x}-1 and g'(x)=3a-2sin x. Notice that e^{x}+1 > 1, therefore, frac {1}{e^{x}+1} in (0,1). Also, -2sin x in [-2,2], therefore, 3a-2sin x in [-2+3a,2+3a]. For the tangent line {{l}_{1}} of the curve f(x)=-e^{x}-x at any point to be always perpendicular to the tangent line {{l}_{2}} of the curve g(x)=3ax+2cos x at some point, the slopes of {{l}_{1}} and {{l}_{2}} must satisfy the condition of perpendicular lines. That is, their product must be equal to -1. Hence, we have the system of inequalities: begin{cases}-2+3aleqslant 0 2+3ageqslant 1end{cases}. Solving this system gives us -frac{1}{3}leqslant aleqslant frac{2}{3}, which is the range of values for a. Therefore, the final answer is boxed{-frac{1}{3}leqslant aleqslant frac{2}{3}}.