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Question Number 209118 Answers: 0 Comments: 0

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Question Number 209099 Answers: 3 Comments: 0

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Question Number 209065 Answers: 1 Comments: 0

The cost of maintaining a school is partly constant and partly varies as the number of students. With 50 students the cost is $15705 and with 40 students the cost is $13305. If the fee per student is $360.00, what is the least number of students for which the school can be run without loss?

$${The}\:{cost}\:{of}\:{maintaining}\:{a}\:{school}\:{is} \\ $$$${partly}\:{constant}\:{and}\:{partly}\:{varies}\:{as} \\ $$$${the}\:{number}\:{of}\:{students}.\:{With}\:\mathrm{50}\:{students} \\ $$$${the}\:{cost}\:{is}\:\$\mathrm{15705}\:{and}\:{with}\:\mathrm{40}\:{students} \\ $$$${the}\:{cost}\:{is}\:\$\mathrm{13305}.\:{If}\:{the}\:{fee}\:{per}\:{student} \\ $$$${is}\:\$\mathrm{360}.\mathrm{00},\:{what}\:{is}\:{the}\:{least}\:{number}\:{of} \\ $$$${students}\:{for}\:{which}\:{the}\:{school}\:{can}\:{be} \\ $$$${run}\:{without}\:{loss}? \\ $$

Question Number 209424 Answers: 1 Comments: 2

Question Number 209062 Answers: 1 Comments: 0

Question Number 209059 Answers: 0 Comments: 3

Compare: 8! and 8!!

$$\mathrm{Compare}: \\ $$$$\mathrm{8}!\:\:\:\mathrm{and}\:\:\:\mathrm{8}!! \\ $$

Question Number 209055 Answers: 0 Comments: 5

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Question Number 209036 Answers: 1 Comments: 1

There are 5 people in a school, 2 boys and 3girls, two students, are chosen at random, find the probability of choosing two different sexes.

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Question Number 209021 Answers: 1 Comments: 7

Hello, I present to you an interesting combinatorics question: A group of people from k families should be seated around a round table, with a_{i} number of people in the i family. Each family member must sit together (i.e. no family member can sit between other family members). There are l spaces around the table. There are seats (l>k). How many ways can we seat k number of families around a round table under these conditions.

$$ \\ $$Hello, I present to you an interesting combinatorics question: A group of people from k families should be seated around a round table, with a_{i} number of people in the i family. Each family member must sit together (i.e. no family member can sit between other family members). There are l spaces around the table. There are seats (l>k). How many ways can we seat k number of families around a round table under these conditions.

Question Number 209016 Answers: 2 Comments: 0

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Question Number 208999 Answers: 3 Comments: 0

Question Number 208980 Answers: 1 Comments: 0

please . find 2^(11001^(666) ) mod 23 thanks.

$${please}\:.\:\:\:\:\:{find}\:\:\mathrm{2}^{\mathrm{11001}^{\mathrm{666}} } {mod}\:\mathrm{23}\:\:\:\:\:\:\:\:{thanks}. \\ $$

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