answer:To find the probability that the selected ticket bears a number which is a multiple of 3, we first need to identify all the numbers between 1 and 24 that are multiples of 3. The multiples of 3 between 1 and 24 are: 3, 6, 9, 12, 15, 18, 21, and 24. There are 8 such numbers. Since there are 24 tickets in total, the probability of selecting a ticket that is a multiple of 3 is the number of favorable outcomes (tickets that are multiples of 3) divided by the total number of possible outcomes (total tickets). So, the probability P is: P = Number of multiples of 3 / Total number of tickets P = 8 / 24 P = 1 / 3 Therefore, the probability that the selected ticket bears a number which is a multiple of 3 is boxed{1/3} .

answer:Let's denote the first discount percentage as ( x )%. This means that after the first discount, the saree is sold for ( 100 - x )% of its original price. The original price of the saree is Rs. 600. After the first discount, the price becomes: [ 600 times frac{100 - x}{100} ] Then, there is an additional 5% discount on the reduced price. So, the price after the second discount is: [ left( 600 times frac{100 - x}{100} right) times frac{95}{100} ] We know that after both discounts, the final price is Rs. 513. So we can set up the equation: [ left( 600 times frac{100 - x}{100} right) times frac{95}{100} = 513 ] Now, we can solve for ( x ): [ 600 times frac{100 - x}{100} times frac{95}{100} = 513 ] [ 600 times (100 - x) times 0.95 = 513 times 100 ] [ 600 times 0.95 times (100 - x) = 51300 ] [ 570 times (100 - x) = 51300 ] [ 100 - x = frac{51300}{570} ] [ 100 - x = 90 ] Now, we can find the value of ( x ): [ x = 100 - 90 ] [ x = 10 ] So, the first discount percentage was boxed{10%} .

answer:By Vieta’s formulas, the average of the sum of the roots equals frac{4}{4} = 1, placing the center of the parallelogram at 1. To shift the center to the origin, let w = z - 1. We substitute z = w + 1 in the given polynomial: [ (w+1)^4 - 4(w+1)^3 + 10b(w+1)^2 - 2(3b^2 + 2b - 2)(w+1) + 4 = 0. ] Expanding and combining like terms, we get: ( w^4 + 4w^3 + (6 + 10b)w^2 + (4 + 20b - 6b^2 - 4b + 4)w + (1 - 4 + 10b - 2(3b^2 + 2b - 2) + 4) = 0. ) Further simplification and assimilation gives: [ w^4 + 4w^3 + (6 + 10b)w^2 + (24b - 6b^2)w + 15b - 6b^2 + 1 = 0. ] The roots are symmetric about the origin: - w_1, -w_1, w_2, -w_2. Therefore, coefficient of w (24b - 6b^2) should be zero: [ 24b - 6b^2 = 0 ] [ 6b(4 - b) = 0 ] [ b = 0 text{ or } b = 4. ] - Check each value of b to determine if the roots form a parallelogram: - For b = 0, the polynomial simplifies and becomes a real root function disproving the parallelogram condition. - For b = 4, substituting and simplifying, it is found that the roots can match the needed conditions for forming a parallelogram. Conclusively, the only value of b that allows the roots to form a parallelogram is boxed{4}.

answer:1. **Analyze the Given Problem**: The problem states that a child has placed four identical cubes such that the letters on the sides of the cubes facing the child form his/her name. We are to determine how the remaining letters are placed on the cube's net and identify the child’s name. 2. **Examine the Configuration**: Referring to the provided image `рис. 6.4a`, we note the positions of the letters 'Н', 'И', 'К', 'А'. 3. **Determine the Initial Steps**: Let us look at the first cube from the right side and draw the letters on the net of this cube. We assume the letter 'А' on the front face and 'И' on the top face. These placements should be correct as they follow the image provided. 4. **Draw on the Net**: - For the first cube, the letters 'А' and 'И' are placed as shown in `рис. 6.4б`. - Now, consider the second cube. Place 'К' and 'Т'. This step aligns with `рис. 6.4в`. 5. **Identify the Missing Piece**: The third cube should logically complete the trio with the final letter 'Н'. Checking by reviewing `рис. 6.4a`, we find that this has been done. 6. **Verify Consistency**: Ensure the fourth cube also matches the pattern we have identified. It confirms our arrangement. 7. **Conclusion from the Drawings**: With the complete net filled with the letters, we determine the letters on the sides facing towards the child which are 'Н', 'И', 'К', and 'А'. 8. **Determine the Child’s Name**: From the letters 'Н', 'И', 'К', 'А', it is safe to conclude that the child’s name is Ника (Nika). So, the final name of the child is: [ boxed{text{Ника}} ] Let's conclude by noting: The arrangement of the letters on the sides corresponds precisely to solve for the name and validate our illustrations.