answer:Solution: (Ⅰ) In triangle ABC, given (2a-c)cos B=bcos C, by the Law of Sines, we have (2sin A-sin C)cos B=sin Bcos C; therefore 2sin Acos B=sin Ccos B+sin Bcos C=sin (B+C)=sin A. Since 0 < A < pi, therefore sin Aneq 0, therefore cos B= frac {1}{2}, and 0 < B < pi, therefore B= frac {pi}{3}; (Ⅱ) Given a=2, c=3, by the Law of Cosines, we get: b^{2}=a^{2}+c^{2}-2accos B=2^{2}+3^{2}-2×2×3cos frac {pi}{3}=7, therefore b= sqrt {7}; then by the Law of Sines, sin C= frac {csin B}{b}= frac {3×sin frac {pi}{3}}{ sqrt {7}}= frac {3 sqrt {21}}{14}.Thus, the answers are: (Ⅰ) B= boxed{frac {pi}{3}}; (Ⅱ) sin C= boxed{frac {3 sqrt {21}}{14}}.

answer: Step-by-Step Solution# Part (1): Total Ways to Select 3 ProductsTo find the total number of ways to select 3 products out of 12, we use the combination formula, which is given by C(n, k) = frac{n!}{k!(n-k)!}, where n is the total number of items, and k is the number of items to choose.For our case, n=12 and k=3, so we have:[C(12, 3) = frac{12!}{3!(12-3)!} = frac{12!}{3!9!} = 220]Therefore, there are boxed{220} different ways to select 3 products from the 12 products.# Part (2): Selecting Exactly 1 Defective ProductTo select exactly 1 defective product and 2 genuine products, we calculate the number of ways to select 1 out of 2 defective products and 2 out of 10 genuine products. This can be done using the combination formula for each selection and then multiplying the results:[C(2, 1) times C(10, 2) = frac{2!}{1!(2-1)!} times frac{10!}{2!(10-2)!}][= 2 times frac{10!}{2!8!} = 2 times 45 = 90]So, there are boxed{90} different ways to select 3 products with exactly 1 defective product.# Part (3): Selecting At Least 1 Defective ProductTo find the number of ways to select at least 1 defective product, we subtract the number of ways to select 3 genuine products (no defective products) from the total number of ways to select 3 products.The number of ways to select 3 genuine products is:[C(10, 3) = frac{10!}{3!(10-3)!} = frac{10!}{3!7!} = 120]Thus, the number of ways to select at least 1 defective product is:[C(12, 3) - C(10, 3) = 220 - 120 = 100]Therefore, there are boxed{100} different ways to select 3 products with at least 1 defective product.

answer:Given the function f(x)=asin omega x-cos omega x, we can start by expressing it in a different form using trigonometric identities. Here's a step-by-step explanation based on the provided solution:1. Express f(x) using a trigonometric identity: We know that a function of the form Asin x + Bcos x can be rewritten as sqrt{A^2+B^2}sin(x+varphi), where tanvarphi = frac{B}{A}. For f(x)=asin omega x-cos omega x, we have sqrt{a^2+1}sin(omega x-varphi), where tan varphi = frac{-1}{a}. - From tan varphi = frac{-1}{a}, solving for a when varphi = frac{pi}{6} because tan(frac{pi}{6}) = frac{1}{sqrt{3}}. Thus, tan varphi = frac{1}{sqrt{3}} implies a = sqrt{3}.2. Find the maximum value: The maximum value of a sine function sin(x) is 1. Therefore, the maximum value of f(x) is sqrt{a^2+1}. - Given that the maximum value of f(x) is 2, we solve sqrt{a^2+1} = 2. Squaring both sides gives a^2 + 1 = 4, simplifying to a^2 = 3, and taking the square root gives a = sqrt{3} (we discard a = -sqrt{3} because a > 0).3. Determine omega: For the function to have a line of symmetry at x = frac{pi}{m}, we consider the periodicity and symmetry of the sine function. - The condition ωx-frac{π}{6}=kπ+frac{π}{2} leads us to frac{ω}{m}π=kπ+frac{2π}{3}, implying that ω = m(k + frac{2}{3}). Since ω and m are positive integers, the smallest value for ω is when m(k + frac{2}{3}) = 2, giving ω = 2 as the smallest integer value satisfying the conditions.4. Count the maximum values within (0,10): The function reaches its maximum when 2x-frac{π}{6}=frac{π}{2}+2kπ. Solving for x gives x = frac{π}{3} + kπ. By testing values of k within the interval (0, 10), we find that k = 0, 1, and 2 give valid x values within the interval, but k = 3 gives x = frac{10π}{3} which is not within (0,10).Therefore, the function f(x) achieves its maximum value 3 times within the interval (0,10).Hence, the answers are: a=boxed{sqrt{3}} and the function f(x) achieves its maximum value boxed{3} times within the interval (0,10).

answer:Let's denote the number of teachers under 40 in the sample as x. To find x, we need to maintain the same ratio of teachers under 40 in the sample as in the total population. The ratio of teachers under 40 in the total teacher population is frac{350}{490}. We want to preserve this ratio in our sample of 70 teachers, so we set up the following proportion:frac{x}{70} = frac{350}{490}Now, we solve for x:begin{align*}frac{x}{70} &= frac{350}{490}x &= frac{350}{490} times 70x &= frac{350 times 70}{490}end{align*}Since 350 = 490 - 140, we can rewrite the above equation as:begin{align*}x &= frac{(490 - 140) times 70}{490}x &= 70 - frac{140 times 70}{490}x &= 70 - frac{140}{7}x &= 70 - 20x &= 50end{align*}Thus, the number of teachers under the age of 40 that should be selected for the sample is boxed{50}.

answer:## Solution.Let the given expression be equal to m. Instead of log _{2}left(3^{n}+7right)=m, we can write3^{n}+7=2^{m}(*) 1 pointThe expression on the left side of this equation gives a remainder of 1 when divided by 3.Powers of the number 2 give remainders of 2 and 1 alternately when divided by 3. The remainder when the number 2^{m} is divided by 3 is equal to 1 if and only if the number m is even.Therefore, m must be an even number.Let m=2 m_{1}. Then 3^{n}+7=4^{m_{1}}.The number on the right side of this equation is divisible by 4.The number 3^{n} gives remainders of 3 and 1 alternately when divided by 4,and since we are adding 7 to this number, we conclude that n must be an even number.Let n=2 n_{1}.We need to determine the natural numbers m_{1}, n_{1} such that 3^{2 n_{1}}+7=2^{2 m_{1}}, or equivalently,left(2^{m_{1}}-3^{n_{1}}right)left(2^{m_{1}}+3^{n_{1}}right)=7The expression in the first parenthesis is smaller than the expression in the second parenthesis.The expression in the second parenthesis is clearly positive, so the expression in the first parenthesis must also be positive.Since 7 is a prime number, the only possibility is2^{m_{1}}-3^{n_{1}}=1, quad 2^{m_{1}}+3^{n_{1}}=7Adding these equations gives 2 cdot 2^{m_{1}}=8, so m_{1}=2,and subtracting them gives 2 cdot 3^{n_{1}}=6, so n_{1}=1.The solution to the equation ( * ) is m=4, n=2.The only natural number n for which the given expression is a natural number is n=2.

answer:(I) Since f(x)=sin frac{x}{2}cos frac{x}{2}+sin ^{2} frac{x}{2},we have f( frac{π}{3})=sin frac{π}{6}cos frac{π}{6}+sin ^{2} frac{π}{6} = frac{1}{2}times frac{ sqrt{3}}{2}+( frac{1}{2})^{2} = frac{1+ sqrt{3}}{4}.(II) f(x)=sin frac{x}{2}cos frac{x}{2}+sin ^{2} frac{x}{2}Simplifying we get: f(x)= frac{1}{2}sin x+ frac{1-cos x}{2} = frac{1}{2}(sin x-cos x)+ frac{1}{2} = frac{ sqrt{2}}{2}sin (x- frac{π}{4})+ frac{1}{2},Given xin(- frac{π}{3}, frac{π}{2}], we have x- frac{π}{4} in (- frac{7π}{12}, frac{π}{4}],Thus, -1leqslant sin (x- frac{π}{4})leqslant frac{ sqrt{2}}{2},This leads to: frac{1- sqrt{2}}{2}leqslant frac{ sqrt{2}}{2}sin (x- frac{π}{4})+ frac{1}{2}leqslant 1,Hence, the range of f(x) is boxed{[frac{1- sqrt{2}}{2},1]}.