answer:Let x-1=0, solving this gives x=1. Therefore, when x=1, the function f(1)=a^{0}+4=5. This means the graph of the function always passes through a fixed point (1,5). Hence, the answer is: boxed{(1,5)}. The function always passes through a fixed point, which is independent of a. According to the problem, we set x-1=0, solve for x=1, and then substitute it back into the function to find the value of f(x). This allows us to determine the coordinates of the fixed point. This question examines the concept of the graph of an exponential function passing through a fixed point (0,1), i.e., setting the exponent to zero to find the corresponding x and y, which are the coordinates of the fixed point the given function passes through. This is considered a basic question.

answer:From the problem, we know that the results of three shots are simulated, and the following 10 groups of random numbers were generated through random simulation. Among these 10 groups of random numbers, those representing exactly two hits in three shots are: 907, 925, 683, totaling 3 groups of random numbers. Therefore, the sought probability is 0.30. Hence, the correct choice is boxed{B}. Analysis From the problem, we know that the results of three shots are simulated, and the following 10 groups of random numbers were generated through random simulation. Among these 10 groups of random numbers, those representing exactly two hits in three shots can be obtained by listing, totaling 3 groups of random numbers. According to the probability formula, the result is obtained.

answer:1. Let ( d_1, d_2, d_3 ) be distinct positive divisors of ( n ) such that ( d_1 < d_2 < d_3 ). Then, we have: [ d_1 + d_2 + d_3 = n ] Since ( d_1, d_2, d_3 ) are distinct divisors of ( n ), it follows that ( d_3 < n ). 2. Given ( d_1 < d_2 < d_3 ), we can infer: [ d_1 + d_2 + d_3 < 3d_3 ] Therefore: [ n < 3d_3 ] Since ( d_3 < n ), we can write: [ d_3 < n < 3d_3 ] This implies: [ n = 2d_3 ] Hence, we have: [ d_1 + d_2 + d_3 = 2d_3 ] Simplifying, we get: [ d_1 + d_2 = d_3 ] 3. Next, we consider the inequality: [ 4d_2 > 2(d_1 + d_2) = d_1 + d_2 + d_3 = n > 2d_2 ] This implies: [ n = 3d_2 ] Therefore: [ 3d_2 = 2d_3 = 2(d_1 + d_2) ] Simplifying, we get: [ 3d_2 = 2d_1 + 2d_2 ] Subtracting ( 2d_2 ) from both sides, we obtain: [ d_2 = 2d_1 ] Substituting ( d_2 = 2d_1 ) into ( d_3 = d_1 + d_2 ), we get: [ d_3 = d_1 + 2d_1 = 3d_1 ] Therefore: [ n = 6d_1 ] 4. For every ( n = 6k ), we can choose ( d_1 = k ), ( d_2 = 2k ), and ( d_3 = 3k ). These values satisfy the conditions ( d_1 + d_2 + d_3 = n ) and ( d_1, d_2, d_3 ) are distinct positive divisors of ( n ). The final answer is ( boxed{ n = 6k } ) for any positive integer ( k ).

answer:1. **Identify the roots and their reciprocals:** We start with the polynomial equation: [ x^3 - 9x^2 + 8x + 2 = 0 ] which has three real roots ( p, q, r ). 2. **Polynomial with reciprocal roots:** If ( p, q, r ) are the roots of the polynomial, their reciprocals ( frac{1}{p}, frac{1}{q}, frac{1}{r} ) will be the roots of the polynomial obtained by reversing the coefficients of ( x^n ). To find the polynomial equation whose roots are the reciprocals of ( p, q, r ), we start by writing down the general transformation: [ a_0 x^n + a_1 x^{n-1} + ldots + a_{n-1} x + a_n = 0 quad Rightarrow quad a_n x^n + a_{n-1} x^{n-1} + ldots + a_1 x + a_0 = 0 ] Given our equation: [ x^3 - 9x^2 + 8x + 2 = 0, ] reversing the coefficients we get: [ 2x^3 + 8x^2 - 9x + 1 = 0. ] Thus, (frac{1}{p}, frac{1}{q}, frac{1}{r}) are the roots of: [ 2x^3 + 8x^2 - 9x + 1 = 0. ] 3. **Use Vieta’s formulas:** Vieta's formulas applied to the polynomial ( 2x^3 + 8x^2 - 9x + 1 = 0 ) tell us: [ frac{1}{p} + frac{1}{q} + frac{1}{r} = -frac{a_{n-1}}{a_n} = -frac{8}{2} = -4, ] [ frac{1}{pq} + frac{1}{pr} + frac{1}{qr} = frac{a_{n-2}}{a_n} = frac{-9}{2}. ] 4. **Calculate the desired expression:** We want to find ( frac{1}{p^2} + frac{1}{q^2} + frac{1}{r^2} ). Using the identity: [ left( frac{1}{p} + frac{1}{q} + frac{1}{r} right)^2 = frac{1}{p^2} + frac{1}{q^2} + frac{1}{r^2} + 2 left( frac{1}{pq} + frac{1}{pr} + frac{1}{qr} right), ] we substitute the known values: [ (-4)^2 = frac{1}{p^2} + frac{1}{q^2} + frac{1}{r^2} + 2 left( -frac{9}{2} right), ] [ 16 = frac{1}{p^2} + frac{1}{q^2} + frac{1}{r^2} - 9, ] [ 16 + 9 = frac{1}{p^2} + frac{1}{q^2} + frac{1}{r^2}, ] [ 25 = frac{1}{p^2} + frac{1}{q^2} + frac{1}{r^2}. ] # Conclusion: [ boxed{25} ]