Let's determine for which vitamins Robert meets at least 75% of his recommended daily allowance (RDA) based on his lunch consisting of a salmon fillet, boiled green beans, and sliced strawberries.
Given information:
- Robert's RDA:
- Vitamin B12: [tex]\(2.4 \, \mu g\)[/tex]
- Vitamin C: [tex]\(75 \, mg\)[/tex]
- Vitamin E: [tex]\(15 \, mg\)[/tex]
Nutrient values of the meal components:
- Salmon fillet:
- Vitamin B12: [tex]\(5.87 \, \mu g\)[/tex]
- Vitamin C: [tex]\(0.0 \, mg\)[/tex]
- Vitamin E: [tex]\(0.60 \, mg\)[/tex]
- 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]
- 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]
Calculate total intake from the meal:
- Total Vitamin B12: [tex]\(5.87 \, \mu g + 0.0 \, \mu g + 0.0 \, \mu g = 5.87 \, \mu g\)[/tex]
- Total Vitamin C: [tex]\(0.0 \, mg + 12.1 \, mg + 49.0 \, mg = 61.1 \, mg\)[/tex]
- Total Vitamin E: [tex]\(0.60 \, mg + 0.57 \, mg + 0.42 \, mg = 1.59 \, mg\)[/tex]
Calculate the percentage of RDA met:
- Percentage of Vitamin B12: [tex]\(\frac{5.87 \, \mu g}{2.4 \, \mu g} \times 100 = 244.58 \%\)[/tex]
- Percentage of Vitamin C: [tex]\(\frac{61.1 \, mg}{75 \, mg} \times 100 = 81.47 \%\)[/tex]
- Percentage of Vitamin E: [tex]\(\frac{1.59 \, mg}{15 \, mg} \times 100 = 10.60 \%\)[/tex]
Determine which vitamins meet at least 75% of RDA:
- Vitamin B12: [tex]\(244.58 \% \geq 75\%\)[/tex]
- Vitamin C: [tex]\(81.47 \% \geq 75\%\)[/tex]
- Vitamin E: [tex]\(10.60 \% < 75\%\)[/tex]
Based on the calculations, Robert meets at least 75% of his RDA for Vitamins B12 and C only.
So, the correct answer is:
B. vitamin B12 and vitamin C only
