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To determine how many hours Maria runs each Saturday, let's break it down step-by-step.
1. First, we start with the total time Maria exercises each Saturday, which is [tex]\( 1 \frac{5}{6} \)[/tex] hours. We need to convert this mixed fraction into an improper fraction.
[tex]\( 1 \frac{5}{6} = 1 + \frac{5}{6} = \frac{6}{6} + \frac{5}{6} = \frac{11}{6} \)[/tex] hours.
2. Next, it is given that Maria runs for [tex]\( \frac{3}{4} \)[/tex] of this total exercise time. We need to calculate [tex]\( \frac{3}{4} \)[/tex] of [tex]\( \frac{11}{6} \)[/tex].
To do this, we'll multiply the fractions:
[tex]\[ \frac{3}{4} \times \frac{11}{6} = \frac{3 \times 11}{4 \times 6} = \frac{33}{24} \][/tex]
3. Simplify the fraction [tex]\( \frac{33}{24} \)[/tex]:
[tex]\[ \frac{33}{24} = \frac{33 \div 3}{24 \div 3} = \frac{11}{8} \][/tex]
Now, convert this improper fraction back to a mixed number:
[tex]\[ \frac{11}{8} = 1 \frac{3}{8} \][/tex]
4. Therefore, the number of hours Maria runs each Saturday is [tex]\( 1 \frac{3}{8} \)[/tex].
Thus, the correct option is:
C [tex]\( 1 \frac{3}{8} \)[/tex]
1. First, we start with the total time Maria exercises each Saturday, which is [tex]\( 1 \frac{5}{6} \)[/tex] hours. We need to convert this mixed fraction into an improper fraction.
[tex]\( 1 \frac{5}{6} = 1 + \frac{5}{6} = \frac{6}{6} + \frac{5}{6} = \frac{11}{6} \)[/tex] hours.
2. Next, it is given that Maria runs for [tex]\( \frac{3}{4} \)[/tex] of this total exercise time. We need to calculate [tex]\( \frac{3}{4} \)[/tex] of [tex]\( \frac{11}{6} \)[/tex].
To do this, we'll multiply the fractions:
[tex]\[ \frac{3}{4} \times \frac{11}{6} = \frac{3 \times 11}{4 \times 6} = \frac{33}{24} \][/tex]
3. Simplify the fraction [tex]\( \frac{33}{24} \)[/tex]:
[tex]\[ \frac{33}{24} = \frac{33 \div 3}{24 \div 3} = \frac{11}{8} \][/tex]
Now, convert this improper fraction back to a mixed number:
[tex]\[ \frac{11}{8} = 1 \frac{3}{8} \][/tex]
4. Therefore, the number of hours Maria runs each Saturday is [tex]\( 1 \frac{3}{8} \)[/tex].
Thus, the correct option is:
C [tex]\( 1 \frac{3}{8} \)[/tex]
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