answer:Let's denote the interest rate of the second investment as ( r ) (in percentage). The simple interest from the first investment at 4% for 7 years is calculated as: [ text{Interest}_1 = P times r_1 times t ] where ( P ) is the principal amount, ( r_1 ) is the interest rate (in decimal form), and ( t ) is the time in years. For the first investment: [ text{Interest}_1 = 900 times 0.04 times 7 ] The simple interest from the second investment at rate ( r % ) for 7 years is: [ text{Interest}_2 = P times r_2 times t ] where ( r_2 ) is the interest rate of the second investment (in decimal form). For the second investment: [ text{Interest}_2 = 900 times frac{r}{100} times 7 ] According to the problem, the interest from the second investment exceeds that from the first by 31.50: [ text{Interest}_2 - text{Interest}_1 = 31.50 ] Substituting the expressions for ( text{Interest}_1 ) and ( text{Interest}_2 ) we get: [ 900 times frac{r}{100} times 7 - 900 times 0.04 times 7 = 31.50 ] Simplifying the equation: [ 6300 times frac{r}{100} - 6300 times 0.04 = 31.50 ] [ 63r - 63 times 4 = 31.50 ] [ 63r - 252 = 31.50 ] Adding 252 to both sides: [ 63r = 31.50 + 252 ] [ 63r = 283.50 ] Dividing both sides by 63: [ r = frac{283.50}{63} ] [ r = 4.5 ] Therefore, the interest rate of the second investment is boxed{4.5%} .

answer:(a) Finding the maximum spread of the wandering point: 1. The point starts at the origin and makes ( a ) steps to the right and ( b ) steps to the left. 2. Given ( a > b ), the net distance from the origin is ( a - b ) to the right. 3. The maximum coordinate that the point can achieve is if it makes all its ( a ) steps to the right first. So the maximum coordinate is ( a ). 4. The minimum coordinate that the point can achieve could be the sum of all left steps first. Thus, the minimum coordinate is ( -b ) (in a theoretical perspective). However, for actual considered positions in the wandering, the effective minimum will be observed as it never retries to start. In any scenario, the lowest reachable from any place can't exceed ( x ) below zero for stability: [ text{Spread} (Delta x) = x+b = a ] The effective spread of the travel thus can't go below a - given bounds constraints always smaller in both via self-checks. 5. So, the maximum possible spread of this wandering point is between ( max (0, a - b) ) = ( a ): [ boxed{a} ] (b) Finding the minimum spread of the wandering point: 1. By assumption, it is let to be ( le 0 ), never exceeding view perspectives such stated ( b to a-b ) below feasible ( x ) and hence: 2. The minimum spread of ( x-y = a ): Computing from basic goal stated each alone: [ text{Spread} (Delta ) = a -y = text{ minimum coordinate } - b = a-b ] 3. So, values bounded as working through become: boxed{a-b} (c) Counting the number of different sequences leading to the maximum spread: 1. From ((a)), to achieve the maximum spread ( a ), all right steps must be taken both one after immediate the other. Rest accordingly at the left: 2. The first step moving right can take from ordered following (below calculation) — giving allowed rotations: 3. However, - Firstly go through b times left then any with above achieving each count feasibly ending; minimum coordinate followings [ ordered starting each first step instant ( a+1; l^{a} ) =( a.b)-Index evaluated So, : a+b sequences only boxed{ a+1 } of unique configuration from traced: boxed{b+1} [ Thus closure at end balances and given queried below remains self-checked: (boxed)

answer:To find the minimal possible value of ( M ), the greatest difference of the numbers in neighboring cells, we need to consider the movement of the rook on an ( n times n ) board. The rook can move horizontally or vertically to an adjacent cell. We aim to minimize the maximum difference between the numbers in these adjacent cells. 1. **Zig-Zag Pattern:** Consider a zig-zag pattern where the rook travels through the board. For an ( n times n ) board, the rook starts at the top-left corner and moves right until it reaches the end of the row, then moves down to the next row and moves left, and so on. This pattern ensures that each cell is visited exactly once. 2. **Numbering the Cells:** Number the cells in the order the rook visits them. For example, in a ( 3 times 3 ) board, the numbering would be: [ begin{array}{ccc} 1 & 2 & 3 6 & 5 & 4 7 & 8 & 9 end{array} ] 3. **Calculating Differences:** In this pattern, the difference between numbers in neighboring cells is either 1 (horizontal movement) or ( 2n-1 ) (vertical movement between rows). For example, in the ( 3 times 3 ) board: - Horizontal differences: ( |2-1| = 1 ), ( |3-2| = 1 ), ( |5-6| = 1 ), etc. - Vertical differences: ( |6-3| = 3 ), ( |7-6| = 1 ), etc. 4. **Generalizing for ( n times n ) Board:** For an ( n times n ) board, the maximum difference in the zig-zag pattern occurs when moving from the end of one row to the start of the next row. This difference is ( 2n-1 ). 5. **Proof of Minimality:** To prove that ( 2n-1 ) is the minimal possible value of ( M ), consider any other pattern of movement. Any pattern must visit each cell exactly once, and the maximum difference will always be at least ( 2n-1 ) due to the need to transition between rows. Therefore, the zig-zag pattern provides the minimal possible value of ( M ). (blacksquare) The final answer is ( boxed{ 2n-1 } ).

answer:Solution: According to the problem, the number of meters of cloth the woman weaves forms an arithmetic sequence, with the first term being a_{1}=5 and the 30th term being a_{30}=1. Therefore, S_{30}= dfrac{30 times (5+1)}{2}=90. Hence, the answer is: boxed{90} meters. By establishing an arithmetic sequence model based on the conditions, we can solve the problem. This question mainly tests the application of arithmetic sequences. Establishing an arithmetic sequence model according to the conditions is the key to solving this problem.