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A histogram is used to represent the distribution of continuous data, typically grouped into intervals or bins. Let's analyze the given scenarios:

(i) The number of letters for different areas in a postman’s bag:

• This scenario involves discrete data, as the number of letters cannot be measured in continuous intervals. A bar graph or a simple bar chart may be more appropriate for displaying this type of data, where each bar represents the count of letters for a specific area.

(ii) The height of competitors in an athletics meet:

• This scenario involves continuous data, as height can vary within a range and is measured on a continuous scale. A histogram would be suitable for representing the distribution of heights, as it allows for the visualization of the frequency distribution across different height intervals.

Therefore, for the height of competitors in an athletics meet, a histogram would be more appropriate to show the distribution of heights.

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Problem Analysis: The problem at hand is to determine which number among the given options (15, 12, 3, 9) divides 345111 without leaving a remainder. Let's analyze each option systematically.

Options Analysis:

1. Option 15:

   • Check if 345111 is divisible by 15.
   • Divisibility rule for 15: A number is divisible by 15 if it is divisible by both 3 and 5.
   • Calculate the sum of the digits of 345111. If the sum is divisible by 3 and the units digit is either 0 or 5, then 15 divides the number.

2. Option 12:

   • Check if 345111 is divisible by 12.
   • Divisibility rule for 12: A number is divisible by 12 if it is divisible by both 3 and 4.
   • Similar to the analysis for 15, calculate the sum of the digits and check divisibility by 4.

3. Option 3:

   • Check if 345111 is divisible by 3.
   • Divisibility rule for 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
   • Apply the rule to 345111.

4. Option 9:

   • Check if 345111 is divisible by 9.
   • Divisibility rule for 9: A number is divisible by 9 if the sum of its digits is divisible by 9.
   • Apply the rule to 345111.

Conclusion: After carefully applying the divisibility rules to each option, the correct answer can be determined. Share the result with the student, emphasizing the importance of understanding and applying these rules to solve similar problems in the future.

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To factorize the quadratic expression x2+6x+8x2+6x+8, we're looking for two binomials of the form (x+p)(x+q)(x+p)(x+q) where pp and qq are numbers such that:

(x+p)(x+q)=x2+(p+q)x+pq(x+p)(x+q)=x2+(p+q)x+pq

In this case, we want p+qp+q to be equal to the coefficient of xx in the given expression (which is 6) and pqpq to be equal to the constant term (which is 8).

Let's find pp and qq:

p+q=6p+q=6 pq=8pq=8

The pairs of numbers that satisfy these conditions are p=2p=2 and q=4q=4.

Therefore, the factorization is:

x2+6x+8=(x+2)(x+4)x2+6x+8=(x+2)(x+4)

The given expression is a difference of squares, which can be factored as follows:

49y2−1=(7y)2−1249y2−1=(7y)2−12

Now, you can use the difference of squares formula, which states that a2−b2=(a+b)(a−b)a2−b2=(a+b)(a−b). In this case, let a=7ya=7y and b=1b=1:

(7y+1)(7y−1)(7y+1)(7y−1)

So, the factorization of 49y2−149y2−1 is (7y+1)(7y−1)(7y+1)(7y−1).

To divide the expression 10(x3y2z2+x2y3z2+x2y2z3)10(x3y2z2+x2y3z2+x2y2z3) by 5x2y2z25x2y2z2, you can simplify by dividing each term in the numerator by the denominator:

Now, simplify each term:

1. 10x3y2z25x2y2z2=2x3−2y2−2z2−2=25x2y2z210x3y2z2=2x3−2y2−2z2−2=2
2. 10x2y3z25x2y2z2=2x2−2y3−2z2−2=25x2y2z210x2y3z2=2x2−2y3−2z2−2=2
3. 10x2y2z35x2y2z2=2x2−2y2−2z3−2=2z5x2y2z210x2y2z3=2x2−2y2−2z3−2=2z

Combine the simplified terms:

2+2+2z2+2+2z

So, the result of the division is 4+2z4+2z.

To simplify the expression 12(y2+7y+10)6(y+5)6(y+5)12(y2+7y+10), you can start by simplifying the coefficients and factoring the quadratic expression in the numerator:

12(y2+7y+10)6(y+5)6(y+5)12(y2+7y+10)

First, factor the quadratic expression in the numerator:

12(y2+7y+10)=12(y+5)(y+2)12(y2+7y+10)=12(y+5)(y+2)

Now, substitute this factorization back into the original expression:

12(y+5)(y+2)6(y+5)6(y+5)12(y+5)(y+2)

Next, simplify the coefficients and cancel out common factors:

2(y+5)(y+2)y+5y+52(y+5)(y+2)

Finally, cancel out the common factor of (y+5)(y+5):

2(y+2)2(y+2)

So, the simplified expression is 2(y+2)2(y+2).