Write two expressions where the solution is 19.

Mathematics

2 answers:

sergij07 [2.7K]2 years ago

7 0

S-1=18
x/10=1.9
these have solution 19

liberstina [14]2 years ago

7 0

Answer:10+9=19. 75÷3=19

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At a school with 100? students, 33 were taking? Arabic, 32 ?Bulgarian, and 40 Chinese. 9 students take only? Arabic, 12 take onl

Stels [109]

Answer: The answer is 4 and 32.

Step-by-step explanation: Let "A", "B" and "C" represents the set of students who were taking Arabic, Bulgarian and Chinese respectively.

The, according to the given information, we have

$n(A)=33,~~n(B)=32,~~n(C)=40,~~n(A\cap B)=14.$

Let 'p' represents the number of students who take all the three languages, then

$n(A\cap B\cap C)=p.$

Also,

$n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)=9\\\\\Rightarrow n(A\cap B)+n(A\cap C)=24+p~~~~~~~~~~~~~~(a),\\\\n(A\cap B)+n(B\cap C)=20+p~~~~~~~~~~~~~~(b),\\\\n(B\cap C)+n(A\cap C)=20+p~~~~~~~~~~~~~~~(c).$

From here, we get after subtracting equation(c) from (b) that

$n(A\cap B)=n(A\cap C)=14.$

Therefore,

$p=14+14-24=4,$ and from equation (a), we find

$n(B\cap C)=24-14=10.$

Thus,

$n(A\cap B\cap C)=4$ and

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\\\\Rightarrow n(A\cap B\cap C)=33+32+40-9-12-20+4\\\\\Rightarrow n(A\cap B\cap C)=68.$