A sheet of address labels is 6 3/4 inches long and 4 1/3 inches wide. The length is divided into sections that are 1 1/8 inches long, and the width is divided into sections that are 1 1/12 inches wide. Solve this multi-step problem involving division with fractions. How many sections are there?(1 point) Responses
24
20
12
10
To find the number of sections in the length, we divide the total length by the length of each section:
6 3/4 ÷ 1 1/8 = (27/4) ÷ (9/8)
To divide fractions, we multiply the first fraction by the reciprocal of the second fraction:
(27/4) x (8/9) = 216/36 = 6
So there are 6 sections in the length.
Similarly, we find the number of sections in the width by dividing the total width by the width of each section:
4 1/3 ÷ 1 1/12 = (13/3) ÷ (13/12)
(13/3) x (12/13) = 12/1 = 12
So there are 12 sections in the width.
To find the total number of sections, we multiply the number of sections in the length by the number of sections in the width:
6 x 12 = 72
Therefore, there are 72 sections in total.
None of the given options (24, 20, 12, 10) match the correct answer of 72.
wrong its one of those answers
Apologies for the incorrect response. Let's re-evaluate the calculation.
To find the number of sections in the length, we divide the total length by the length of each section:
6 3/4 ÷ 1 1/8 = (27/4) ÷ (9/8)
To divide fractions, we multiply the first fraction by the reciprocal of the second fraction:
(27/4) x (8/9) = 216/36 = 6.
So there are 6 sections in the length.
Similarly, we find the number of sections in the width by dividing the total width by the width of each section:
4 1/3 ÷ 1 1/12 = (13/3) ÷ (13/12)
(13/3) x (12/13) = 12/1 = 12.
So there are 12 sections in the width.
To find the total number of sections, we multiply the number of sections in the length by the number of sections in the width:
6 x 12 = 72.
Therefore, there are 72 sections in total.
Based on the calculations, none of the given options (24, 20, 12, 10) match the correct answer of 72. It seems the answer choices provided are not correct.