Let's determine whether Robert meets at least 75% of the recommended daily intake for each vitamin based on his lunch components: salmon filet, boiled green beans, and sliced strawberries.
1. Vitamin B12:
- Salmon filet: [tex]\(5.87\, \mu g\)[/tex]
- Boiled green beans: [tex]\(0.0\, \mu g\)[/tex]
- Sliced strawberries: [tex]\(0.0\, \mu g\)[/tex]
Total Vitamin B12 intake:
[tex]\[
5.87\, \mu g + 0.0\, \mu g + 0.0\, \mu g = 5.87\, \mu g
\][/tex]
Percentage of the recommended intake:
[tex]\[
\frac{5.87\, \mu g}{2.4\, \mu g} \times 100 = 244.583\%
\][/tex]
Since 244.583% is greater than 75%, Robert meets the 75% criteria for Vitamin B12.
2. Vitamin C:
- Salmon filet: [tex]\(0.0\, mg\)[/tex]
- Boiled green beans: [tex]\(12.1\, mg\)[/tex]
- Sliced strawberries: [tex]\(49.0\, mg\)[/tex]
Total Vitamin C intake:
[tex]\[
0.0\, mg + 12.1\, mg + 49.0\, mg = 61.1\, mg
\][/tex]
Percentage of the recommended intake:
[tex]\[
\frac{61.1\, mg}{75\, mg} \times 100 = 81.467\%
\][/tex]
Since 81.467% is greater than 75%, Robert meets the 75% criteria for Vitamin C.
3. Vitamin E:
- Salmon filet: [tex]\(0.60\, mg\)[/tex]
- Boiled green beans: [tex]\(0.57\, mg\)[/tex]
- Sliced strawberries: [tex]\(0.42\, mg\)[/tex]
Total Vitamin E intake:
[tex]\[
0.60\, mg + 0.57\, mg + 0.42\, mg = 1.59\, mg
\][/tex]
Percentage of the recommended intake:
[tex]\[
\frac{1.59\, mg}{15\, mg} \times 100 = 10.6\%
\][/tex]
Since 10.6% is less than 75%, Robert does not meet the 75% criteria for Vitamin E.
Based on the calculations, Robert meets at least 75% of the recommended daily intake for Vitamin B12 and Vitamin C only.
Therefore, the correct answer is:
B. vitamin B12 and vitamin C only
