1. Problem 1: Compute the total receipts for Kamukamu and Tweziimbe from the given sales and unit prices. 2. Given data for the four sales records and unit prices. 3. Kamukamu period 1 sold $2520$ litres milk, $35$ bags maize and $10$ bags beans. 4. Tweziimbe period 1 sold $2314$ litres milk, $41$ bags maize and $9$ bags beans. 5. Kamukamu period 2 sold $3254$ litres milk, $42$ bags maize and $8$ bags beans. 6. Tweziimbe period 2 sold $2719$ litres milk, $32$ bags maize and $11$ bags beans. 7. Unit prices are milk per litre $700$, maize per kg $1000$, beans per kg $3500$, and each bag is $120$ kg. 8. Value per maize bag is $120\times1000=120000$ and value per beans bag is $120\times3500=420000$. 9. Compute Kamukamu period 1 revenues componentwise. $$2520\times700=1764000$$ $$35\times120000=4200000$$ $$10\times420000=4200000$$ 10. Sum Kamukamu period 1 revenue. $$1764000+4200000+4200000=10164000$$ 11. Compute Tweziimbe period 1 revenues componentwise. $$2314\times700=1619800$$ $$41\times120000=4920000$$ $$9\times420000=3780000$$ 12. Sum Tweziimbe period 1 revenue. $$1619800+4920000+3780000=10319800$$ 13. Compute Kamukamu period 2 revenues componentwise. $$3254\times700=2277800$$ $$42\times120000=5040000$$ $$8\times420000=3360000$$ 14. Sum Kamukamu period 2 revenue. $$2277800+5040000+3360000=10677800$$ 15. Compute Tweziimbe period 2 revenues componentwise. $$2719\times700=1903300$$ $$32\times120000=3840000$$ $$11\times420000=4620000$$ 16. Sum Tweziimbe period 2 revenue. $$1903300+3840000+4620000=10363300$$ 17. Total receipts for each group (sum of their two records). $$\text{Kamukamu total}=10164000+10677800=20841800$$ $$\text{Tweziimbe total}=10319800+10363300=20683100$$ 18. Conclusion for Problem 1: Kamukamu earned more overall by $$20841800-20683100=158700$$ 1. Problem 2: Decide whether to buy one-size-fits-all T-shirts if at least $70\%$ of classmates fit in at least two sizes. 2. Given counts: $n(M)=100$, $n(L)=100$, $n(S)=76$, $n(S\cap L)=50$, $n(M\cap L)=70$, $n(S\cap M)=60$, $n(\text{none})=4$, class total $140$, and let $x=n(S\cap M\cap L)$. 3. Use inclusion--exclusion for total class size. $$140=n(S)+n(M)+n(L)-[n(S\cap M)+n(S\cap L)+n(M\cap L)]+x+\text{none}$$ 4. Substitute numbers and solve for $x$. $$140=76+100+100-(60+50+70)+x+4$$ $$140=276-180+x+4$$ $$140=100+x$$ $$x=40$$ 5. Number who fit in at least two sizes equals sum of pairwise intersections minus twice the triple intersection. $$\text{at least two}=n(S\cap M)+n(S\cap L)+n(M\cap L)-2x$$ $$=60+50+70-2\times40$$ $$=180-80=100$$ 6. Required threshold $70\%$ of $140$ is $$0.70\times140=98$$ 7. Since $100\ge98$, the condition is satisfied and you should buy and put on T-shirts as a class. 1. Problem 3(a): Arrange purchases in arrays and determine which school spent most in the two weeks. 2. Interpret first-week data with "potatoes" treated as cassava for price availability, then sum per school across two weeks. 3. Totals per school (posho, rice, cassava) across the two weeks are: $$\text{Nalinya}=[5,1,5]$$ $$\text{Ndejje}=[3,4,1]$$ $$\text{Sazza}=[4,1,3]$$ 4. Price vector (posho, rice, cassava): $$[20000,30000,10000]$$ 5. Compute spending as dot product of totals with prices. $$\text{Nalinya spend}=5\times20000+1\times30000+5\times10000=100000+30000+50000=180000$$ $$\text{Ndejje spend}=3\times20000+4\times30000+1\times10000=60000+120000+10000=190000$$ $$\text{Sazza spend}=4\times20000+1\times30000+3\times10000=80000+30000+30000=140000$$ 6. Conclusion for 3(a): Ndejje S.S spent the most with a total of $190000$. 1. Problem 3(b)(i): Arrange survey results in a Venn-style table with counts. 2. Given survey numbers: $n(C)=47$, $n(P)=53$, $n(R)=45$, $P\text{ only}=23$, $C\text{ only}=10$, $R\text{ only}=5$, and $P\cap C\cap R=15$. 3. Let $x=n(P\cap C\text{ only})$, $y=n(P\cap R\text{ only})$, $z=n(C\cap R\text{ only})$ and solve from totals. $$x+y=53-23-15=15$$ $$x+z=47-10-15=22$$ $$y+z=45-5-15=25$$ 4. Solving gives $x=6$, $y=9$, $z=16$. 5. The Venn counts are: $P\text{ only}=23$, $C\text{ only}=10$, $R\text{ only}=5$, $P\cap C\text{ only}=6$, $P\cap R\text{ only}=9$, $C\cap R\text{ only}=16$, and all three $15$. 1. Problem 3(b)(ii): Number preferring at least two foodstuffs equals exactly-two plus all-three. $$\text{exactly two}=x+y+z=6+9+16=31$$ $$\text{at least two}=31+15=46$$ 1. Problem 3(b)(iii): Total students sampled is sum of onlys, exactly-twos and all-three. $$\text{sample size}=23+10+5+31+15=84$$ 1. Problem 3(b)(iv): Chance a random student prefers rice equals $n(R)/\text{sample size}$. $$\text{probability}=45/84=15/28\approx0.5357$$ 2. Conclusion: The probability $15/28$ is greater than $0.5$ so more than half of the sampled students like rice, but posho has the largest raw count $53$, so posho is the most preferred overall in the sample. Final answers summary. 1. Problem 1 final totals: Kamukamu $20841800$, Tweziimbe $20683100$, Kamukamu earned more by $158700$. 2. Problem 2 final decision: Buy the T-shirts since $100$ students fit at least two sizes which meets the $70\%$ requirement of $98$. 3. Problem 3 final results: (a) Ndejje spent most with $190000$. (b)(ii) At least two foods liked by $46$ students, (iii) sample size $84$, (iv) probability a random student prefers rice is $15/28\approx0.5357$ and posho remains the single most preferred item.