A display has 666 packs of marbles with a total mass of 629\,\text{g}629g629, start text, g, end text. The packaging of each pack has a mass of \dfrac{2}{3}\,\text{g} 3 2 gstart fraction, 2, divided by, 3, end fraction, start text, g, end text and each marble has a mass of 4\dfrac{1}{2}\,\text{g}4 2 1 g4, start fraction, 1, divided by, 2, end fraction, start text, g, end text. Which equation can we use to determine mmm, the number of marbles per pack? Choose 1 answer: Choose 1 answer: (Choice A) A 6\cdot 4\dfrac{1}{2}m+\dfrac{2}{3}=6296⋅4 2 1 m+ 3 2 =6296, dot, 4, start fraction, 1, divided by, 2, end fraction, m, plus, start fraction, 2, divided by, 3, end fraction, equals, 629 (Choice B) B 6m\left(4\dfrac{1}{2}+\dfrac{2}{3}\right)=6296m(4 2 1 + 3 2 )=6296, m, left parenthesis, 4, start fraction, 1, divided by, 2, end fraction, plus, start fraction, 2, divided by, 3, end fraction, right parenthesis, equals, 629 (Choice C) C 6m+4\dfrac{1}{2}+\dfrac{2}{3}=6296m+4 2 1 + 3 2 =6296, m, plus, 4, start fraction, 1, divided by, 2, end fraction, plus, start fraction, 2, divided by, 3, end fraction, equals, 629 (Choice D) D 6\left(4\dfrac{1}{2}m+\dfrac{2}{3}\right)=6296(4 2 1 m+ 3 2 )=629
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Answer:
The set of odd numbers less than 7 is written as: {odd numbers less than 7}. ... (ii) The set of all vowels of the English alphabet. Therefore, V = {a, e, i, o, u} → Roster Form. (iii) The set of all odd numbers less than 9.1
Answer:
A display has 666 packs of marbles with a total mass of 629\,\text{g}629g629, start text, g, end text. The packaging of each pack has a mass of \dfrac{2}{3}\,\text{g} 3 2 gstart fraction, 2, divided by, 3, end fraction, start text, g, end text and each marble has a mass of 4\dfrac{1}{2}\,\text{g}4 2 1 g4, start fraction, 1, divided by, 2, end fraction, start text, g, end text. Which equation can we use to determine mmm, the number of marbles per pack? Choose 1 answer: Choose 1 answer: (Choice A) A 6\cdot 4\dfrac{1}{2}m+\dfrac{2}{3}=6296⋅4 2 1 m+ 3 2 =6296, dot, 4, start fraction, 1, divided by, 2, end fraction, m, plus, start fraction, 2, divided by, 3, end fraction, equals, 629 (Choice B) B 6m\left(4\dfrac{1}{2}+\dfrac{2}{3}\right)=6296m(4 2 1 + 3 2 )=6296, m, left parenthesis, 4, start fraction, 1, divided by, 2, end fraction, plus, start fraction, 2, divided by, 3, end fraction, right parenthesis, equals, 629 (Choice C) C 6m+4\dfrac{1}{2}+\dfrac{2}{3}=6296m+4