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Sagot :
To determine for which vitamins Robert met at least 75% of his recommended daily allowance (RDA), we need to compare the total intake from his meal with the recommended amounts and calculate the respective percentages.
Let's gather all the given information:
Recommended Daily Allowances (RDA):
- Vitamin B12: [tex]\(2.4 \mu g\)[/tex] per day
- Vitamin C: [tex]\(75 mg\)[/tex] per day
- Vitamin E: [tex]\(15 mg\)[/tex] per day
Nutrient values in Robert's meal:
1. Salmon fillet (124g):
- Vitamin B12: [tex]\(5.87 \mu g\)[/tex]
- Vitamin C: [tex]\(0.0 mg\)[/tex]
- Vitamin E: [tex]\(0.60 mg\)[/tex]
2. Boiled green beans (1 cup):
- Vitamin B12: [tex]\(0.0 \mu g\)[/tex]
- Vitamin C: [tex]\(12.1 mg\)[/tex]
- Vitamin E: [tex]\(0.57 mg\)[/tex]
3. Strawberries, sliced (1/2 cup):
- Vitamin B12: [tex]\(0.0 \mu g\)[/tex]
- Vitamin C: [tex]\(49.0 mg\)[/tex]
- Vitamin E: [tex]\(0.42 mg\)[/tex]
Step-by-step Solution:
1. Calculate total intake of each vitamin:
- Total Vitamin B12:
[tex]\[ 5.87 \mu g (from salmon) + 0.0 \mu g (from green beans) + 0.0 \mu g (from strawberries) = 5.87 \mu g \][/tex]
- Total Vitamin C:
[tex]\[ 0.0 mg (from salmon) + 12.1 mg (from green beans) + 49.0 mg (from strawberries) = 61.1 mg \][/tex]
- Total Vitamin E:
[tex]\[ 0.60 mg (from salmon) + 0.57 mg (from green beans) + 0.42 mg (from strawberries) = 1.59 mg \][/tex]
2. Calculate the percentage of RDA met for each vitamin:
- Percentage of Vitamin B12:
[tex]\[ \left( \frac{5.87 \mu g}{2.4 \mu g} \right) \times 100 \% = 244.58 \% \][/tex]
- Percentage of Vitamin C:
[tex]\[ \left( \frac{61.1 mg}{75 mg} \right) \times 100 \% = 81.47 \% \][/tex]
- Percentage of Vitamin E:
[tex]\[ \left( \frac{1.59 mg}{15 mg} \right) \times 100 \% = 10.6 \% \][/tex]
3. Determine which vitamins meet at least 75% of RDA:
- Vitamin B12: 244.58% (Met 75% RDA requirement)
- Vitamin C: 81.47% (Met 75% RDA requirement)
- Vitamin E: 10.6% (Did not meet 75% RDA requirement)
Based on the above calculations, Robert met at least 75% of the recommended daily allowance for Vitamin B12 and Vitamin C only.
Correct Answer:
B. Vitamin B12 and Vitamin C only
Let's gather all the given information:
Recommended Daily Allowances (RDA):
- Vitamin B12: [tex]\(2.4 \mu g\)[/tex] per day
- Vitamin C: [tex]\(75 mg\)[/tex] per day
- Vitamin E: [tex]\(15 mg\)[/tex] per day
Nutrient values in Robert's meal:
1. Salmon fillet (124g):
- Vitamin B12: [tex]\(5.87 \mu g\)[/tex]
- Vitamin C: [tex]\(0.0 mg\)[/tex]
- Vitamin E: [tex]\(0.60 mg\)[/tex]
2. Boiled green beans (1 cup):
- Vitamin B12: [tex]\(0.0 \mu g\)[/tex]
- Vitamin C: [tex]\(12.1 mg\)[/tex]
- Vitamin E: [tex]\(0.57 mg\)[/tex]
3. Strawberries, sliced (1/2 cup):
- Vitamin B12: [tex]\(0.0 \mu g\)[/tex]
- Vitamin C: [tex]\(49.0 mg\)[/tex]
- Vitamin E: [tex]\(0.42 mg\)[/tex]
Step-by-step Solution:
1. Calculate total intake of each vitamin:
- Total Vitamin B12:
[tex]\[ 5.87 \mu g (from salmon) + 0.0 \mu g (from green beans) + 0.0 \mu g (from strawberries) = 5.87 \mu g \][/tex]
- Total Vitamin C:
[tex]\[ 0.0 mg (from salmon) + 12.1 mg (from green beans) + 49.0 mg (from strawberries) = 61.1 mg \][/tex]
- Total Vitamin E:
[tex]\[ 0.60 mg (from salmon) + 0.57 mg (from green beans) + 0.42 mg (from strawberries) = 1.59 mg \][/tex]
2. Calculate the percentage of RDA met for each vitamin:
- Percentage of Vitamin B12:
[tex]\[ \left( \frac{5.87 \mu g}{2.4 \mu g} \right) \times 100 \% = 244.58 \% \][/tex]
- Percentage of Vitamin C:
[tex]\[ \left( \frac{61.1 mg}{75 mg} \right) \times 100 \% = 81.47 \% \][/tex]
- Percentage of Vitamin E:
[tex]\[ \left( \frac{1.59 mg}{15 mg} \right) \times 100 \% = 10.6 \% \][/tex]
3. Determine which vitamins meet at least 75% of RDA:
- Vitamin B12: 244.58% (Met 75% RDA requirement)
- Vitamin C: 81.47% (Met 75% RDA requirement)
- Vitamin E: 10.6% (Did not meet 75% RDA requirement)
Based on the above calculations, Robert met at least 75% of the recommended daily allowance for Vitamin B12 and Vitamin C only.
Correct Answer:
B. Vitamin B12 and Vitamin C only
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