1. For the lunch and gift shop problem, state given info: - Students brought packed lunch: 43 - Students visited gift shop: 14 - Students neither brought lunch nor visited gift shop: 22 2. Let the number who brought lunch and visited gift shop be $x$. 3. Then, those who brought lunch but did not visit gift shop are $43 - x$. 4. Those who did not bring lunch but visited gift shop are $14 - x$. 5. Students who neither brought lunch nor visited gift shop = 22 (given). 6. Total students = (brought lunch and visited) + (brought lunch, no visit) + (no lunch, visited) + (neither) = $x + (43-x) + (14-x) + 22 = 43 + 14 + 22 - x = 79 - x$. 7. This total should equal total students, so $79 - x$ is total. 8. Since number can't be negative, $x$ must be such that each branch count is nonnegative. 9. Usually $x$ is given or assumed to be the intersection of lunch and visit counts; use the principle of inclusion-exclusion: $$\text{Total} = |L| + |G| - |L \cap G| + |\text{neither}|\Rightarrow \text{Total} = 43 + 14 - x + 22$$ So $\text{Total} = 79 - x$ as above. 10. The problem does not give total students, so find $x$ consistent with total students: - Sum all must equal total = $T$. - Total equals $79 - x$. 11. The question is: How many students brought packed lunch and did not visit gift shop? That's $43 - x$. 12. Without total $T$ or additional constraints, cannot find exact $x$, but using minimal overlap assumption for maximum $x$, assume $x = 14$ (all who visited gift shop also brought lunch). 13. Then packed lunch and no visit: $43 - 14 = 29$. 14. For the Spanish and French test problem: - Pass Spanish & Pass French = 14 - Pass Spanish & Fail French = 12 - Fail Spanish total = 24, of which Pass French = 9, Fail French = $24 - 9 = 15$ 15. Total students = sum all outcomes = $14 + 12 + 9 + 15 = 50$. 16. Probability student failed at least one test = 1 - probability passed both. 17. Probability passed both tests = $\frac{14}{50} = \frac{7}{25}$. 18. Therefore, probability failed at least one test = $1 - \frac{7}{25} = \frac{18}{25}$. Final answers: - Students who brought packed lunch and did not visit gift shop: 29 - Probability failed at least one test: $\frac{18}{25}$