Answer :
To determine which vitamins Robert met at least 75% of his recommended daily allowance, we need to find the totals of each vitamin in his lunch and then calculate the percentages of the recommended intake these totals represent.
First, let's summarize the nutrient values in Robert's lunch:
- Salmon Fillet:
- Vitamin B12: 5.87 µg
- Vitamin C: 0.0 mg
- Vitamin E: 0.60 mg
- Boiled Green Beans (1 cup):
- Vitamin B12: 0.0 µg
- Vitamin C: 12.1 mg
- Vitamin E: 0.57 mg
- Strawberries, Sliced (1/2 cup):
- Vitamin B12: 0.0 µg
- Vitamin C: 49.0 mg
- Vitamin E: 0.42 mg
Next, we calculate the total intake of each vitamin from this 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]
Now, we calculate the percentages of the recommended daily allowances represented by these totals:
- Percentage of Vitamin B12:
[tex]\[ \left(\frac{5.87 \mu g}{2.4 \mu g}\right) \times 100 \approx 244.58 \% \][/tex]
- Percentage of Vitamin C:
[tex]\[ \left(\frac{61.1 mg}{75 mg}\right) \times 100 \approx 81.47 \% \][/tex]
- Percentage of Vitamin E:
[tex]\[ \left(\frac{1.59 mg}{15 mg}\right) \times 100 \approx 10.6 \% \][/tex]
Finally, we determine which of these percentages are at least 75% of the recommended intake:
- Vitamin B12: [tex]\(244.58\%\)[/tex] which is indeed >= 75%
- Vitamin C: [tex]\(81.47\%\)[/tex] which is also >= 75%
- Vitamin E: [tex]\(10.6\%\)[/tex] which is less than 75%
Therefore, Robert met at least 75% of his recommended daily allowance for Vitamin B12 and Vitamin C only.
The correct answer is:
B. vitamin B12 and vitamin C only
First, let's summarize the nutrient values in Robert's lunch:
- Salmon Fillet:
- Vitamin B12: 5.87 µg
- Vitamin C: 0.0 mg
- Vitamin E: 0.60 mg
- Boiled Green Beans (1 cup):
- Vitamin B12: 0.0 µg
- Vitamin C: 12.1 mg
- Vitamin E: 0.57 mg
- Strawberries, Sliced (1/2 cup):
- Vitamin B12: 0.0 µg
- Vitamin C: 49.0 mg
- Vitamin E: 0.42 mg
Next, we calculate the total intake of each vitamin from this 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]
Now, we calculate the percentages of the recommended daily allowances represented by these totals:
- Percentage of Vitamin B12:
[tex]\[ \left(\frac{5.87 \mu g}{2.4 \mu g}\right) \times 100 \approx 244.58 \% \][/tex]
- Percentage of Vitamin C:
[tex]\[ \left(\frac{61.1 mg}{75 mg}\right) \times 100 \approx 81.47 \% \][/tex]
- Percentage of Vitamin E:
[tex]\[ \left(\frac{1.59 mg}{15 mg}\right) \times 100 \approx 10.6 \% \][/tex]
Finally, we determine which of these percentages are at least 75% of the recommended intake:
- Vitamin B12: [tex]\(244.58\%\)[/tex] which is indeed >= 75%
- Vitamin C: [tex]\(81.47\%\)[/tex] which is also >= 75%
- Vitamin E: [tex]\(10.6\%\)[/tex] which is less than 75%
Therefore, Robert met at least 75% of his recommended daily allowance for Vitamin B12 and Vitamin C only.
The correct answer is:
B. vitamin B12 and vitamin C only
