answer:To calculate the total amount of the investment after 10 years with an interest rate of 4% compounded annually, use the formula: [ A = P(1 + r)^n ] where ( P = 8000 ), ( r = 0.04 ), and ( n = 10 ). Substituting these values into the formula gives: [ A = 8000(1 + 0.04)^{10} ] [ A = 8000(1.04)^{10} ] [ A = 8000 times 1.48024 ] [ A = 11,841.92 ] To the nearest whole number, the total amount of the investment after 10 years is ( boxed{11842} ).

answer:**Answer:** Given the universal set U={1,2,3,4,5}, and C_UP={4,5}, then the set P={1,2,3}. Analyzing the options, we find that: In option A, {xinmathbb{N}^* mid |x|<4}={1,2,3}, which matches the condition; In option B, {xinmathbb{N}^* mid x<6}={1,2,3,4,5}, which does not match the condition; In option C, {xinmathbb{N}^* mid x^2leq16}={1,2,3,4}, which does not match the condition; In option D, {xinmathbb{N}^* mid 1leq xleq4}={1,2,3,4}, which does not match the condition; Therefore, the correct choice is boxed{A}. **Analysis:** According to the problem, by the operation of the complement, we get P={1,2,3}. Analyzing the options in sequence, we find that in option A, {xinmathbb{N}^* mid |x|<4}={1,2,3}, in option B, {xinmathbb{N}^* mid x<6}={1,2,3,4,5}, in option C, {xinmathbb{N}^* mid x^2leq16}={1,2,3,4}, and in option D, {xinmathbb{N}^* mid 1leq xleq4}={1,2,3,4}. Comparing these with the set P, we can determine the answer.

answer:1. **Base Case:** For ( n = 1 ), we define: [ P_{1}(x) = x - 1 ] Clearly, ( P_1(x) ) has coefficients ( 1 ) and (-1), and it is divisible by ( (x - 1)^1 = x - 1 ). 2. **Recursive Definition:** Assume ( P_{n}(x) ) is a polynomial with coefficients ( 0, -1, 1 ) and degree not greater than ( 2^n - 1 ), which is divisible by ( (x - 1)^n ). We define the next polynomial ( P_{n+1}(x) ) as: [ P_{n+1}(x) = left(x^{2^{n}} - 1right)P_{n}(x) ] 3. **Degree Verification:** The degree of ( P_{n+1}(x) ) is: [ operatorname{deg} P_{n+1}(x) = operatorname{deg} left( x^{2^{n}} - 1 right) + operatorname{deg} P_{n}(x) ] Since ( operatorname{deg} (x^{2^n} - 1) = 2^n ) and ( operatorname{deg}P_n(x) = 2^n - 1 ), we have: [ operatorname{deg} P_{n+1}(x) = 2^n + (2^n - 1) = 2^{n+1} - 1 ] This confirms that ( P_{n+1}(x) ) has the correct degree. 4. **Coefficient Verification:** The polynomials ( P_n(x) ) have coefficients ( 0, -1, 1 ). Consider the product ((x^{2^n} - 1)P_n(x)): - ( x^{2^n} ) has coefficient ( 1 ). - ( -1 ) has coefficient (-1). - ( P_n(x) ) has coefficients ( 0, -1, 1 ), ensuring ( P_{n+1}(x) ) only has terms where coefficients are ( 0, -1, 1 ). - Since ( P_n(x) ) has no overlapping terms, the coefficients are preserved in the product. 5. **Divisibility Verification:** Notice that ( x^{2^n} - 1 ) can be factored as: [ x^{2^n} - 1 = (x-1)(x^{2^n-1} + x^{2^n-2} + cdots + 1) ] Thus, ( x^{2^n} - 1 ) is divisible by ( x - 1 ). So, we can write: [ P_{n+1}(x) = (x^{2^n} - 1)P_n(x) ] Given ( P_n(x) ) is divisible by ( (x - 1)^n ), it follows: [ P_{n+1}(x) = (x-1)(Q(x))P_n(x) ] Since both ( (x-1) ) from ( x^{2^n} - 1 ) and ( (x-1)^n ) from ( P_n(x) ) are present, ( P_{n+1}(x) ) is divisible by ( (x-1)^{n+1} ). 6. **Conclusion:** By mathematical induction, we have shown that for any natural number ( n ), there exists a non-zero polynomial ( P(x) ) with coefficients ( 0, -1, 1 ), with degree not exceeding ( 2^n ), that is divisible by ( (x-1)^n ). [ blacksquare ]

answer:Since xi sim B(10, 0.6), we have Exi = 10 times 0.6 = 6, Dxi = 10 times 0.6 times 0.4 = 2.4, Given xi + eta = 8, we get Eeta = E(8 - xi) = 2, Deta = D(8 - xi) = 2.4 Therefore, the correct choice is boxed{text{B}}.