answer:1. **Calculate the cost per pen**: The cost per pen can be calculated by dividing the total cost by the number of pens in a pack. [ text{Cost per pen} = frac{45}{150} = 0.30 text{ per pen} ] 2. **Calculate the total cost for 3600 pens**: Multiply the cost per pen by the total number of pens desired. [ text{Total cost} = 3600 times 0.30 = 1080 ] 3. **Conclusion**: Therefore, buying 3600 pens would cost boxed{1080}.

answer:Let's denote the current ages of the three brothers as follows: - Richard's current age = R - David's current age = D - Scott's current age = S From the information given, we can set up the following equations: 1. Richard is 6 years older than David: R = D + 6 2. David was 12 years old 2 years ago, so his current age is: D = 12 + 2 D = 14 3. In 8 years, Richard will be twice as old as Scott: R + 8 = 2(S + 8) Now we can substitute the value of D into the first equation to find R: R = D + 6 R = 14 + 6 R = 20 Now we have Richard's current age. Let's use this information in the third equation to find Scott's current age: R + 8 = 2(S + 8) 20 + 8 = 2(S + 8) 28 = 2S + 16 28 - 16 = 2S 12 = 2S S = 12 / 2 S = 6 Scott's current age is 6 years. Now we can find out how many years older David is than Scott: D - S = 14 - 6 D - S = 8 David is boxed{8} years older than Scott.

answer:【Analysis】 This question examines the probability calculation of the normal distribution, which is a basic question. By the given information, we find (μ=78), (σ=4). According to (P(66 < X < 90)=P(μ-3σ < X leqslant μ+3σ)), we calculate the value of (P(66 < X < 90)), and then use the symmetry of the normal distribution to solve the problem. 【Solution】 Given the normal distribution (N(78,16)), we have (μ=78), (σ=4), Then (P(66 < X < 90)=P(78-3×4 < X < 78+3×4)=0.997). (∴P(X geqslant 90)=dfrac{1}{2} (1-0.997)=0.0015). Thus, the estimated percentage of students whose test scores are not less than (90) is (0.13%). Therefore, the correct choice is (boxed{text{A}}).

answer:Solution: (Ⅰ) When (a=1), the function becomes (f(x) = 4x^2 - 2bx - 1 + b), Let (f(x) = 0), then (x = frac{1}{2}), or (x = frac{-1+b}{2}), If the function (f(x)) has two different zeros within the domain ([0,1]), then (frac{-1+b}{2} in [0,1]), and (frac{-1+b}{2} neq frac{1}{2}), Solving this gives: (b in [1,2) cup (2,3]) Thus, the range of (b) is boxed{[1,2) cup (2,3]}. Proof: (Ⅱ) To prove: (f(x) + M > 0), which means to prove: (f(x)_{text{max}} + f(x)_{text{min}} > 0) Since the graph of (f(x) = 4ax^2 - 2bx - a + b) is an upward-opening parabola, symmetric about the line (x = frac{b}{4a}), ① If (frac{b}{4a} < 0), or (frac{b}{4a} > 1), then (f(x)_{text{max}} + f(x)_{text{min}} = f(0) + f(1) = -a + b + 3a - b = 2a > 0); ② If (0 leqslant frac{b}{4a} < frac{1}{2}), i.e., (0 leqslant b < 2a), then (f(x)_{text{max}} + f(x)_{text{min}} = f(frac{b}{4a}) + f(1) = -a + b - frac{b^2}{4a} + 3a - b = 2a - frac{b^2}{4a} = frac{8a^2 - b^2}{4a} > frac{4a^2}{4a} = a > 0); ③ If (frac{1}{2} leqslant frac{b}{4a} leqslant 1), i.e., (2a leqslant b leqslant 4a), then (f(x)_{text{max}} + f(x)_{text{min}} = f(frac{b}{4a}) + f(0) = -a + b - frac{b^2}{4a} - a + b = 2b - 2a - frac{b^2}{4a} = frac{8ab - 8a^2 - b^2}{4a} = frac{-(b-4a)^2 + 8a^2}{4a} geqslant frac{4a^2}{4a} = a > 0); Therefore, it is always true that (f(x)_{text{max}} + f(x)_{text{min}} > 0), which means (f(x) + M > 0) is proven.