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3 years ago

Find the percent of change. 120lbs to 1801bs

1 answer:

3 0

Quick Note: I'm pretty sure you meant 180LBS and not 1801bs, so I'm using 180, but if it was 1801 please tell me

Percent of change = $\frac{V2 - V1}{V1} *100$

Substitute 180 for V2 and 160 for V1
Percent of change = $\frac{180-160}{160} * 100$

Solve
Percent of change = $\frac{20}{160} *100$
Percent of change = 0.125 x 100
Percent of change = 12.5%

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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

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

Thus, the number of students who take all the three languages is 4 and the number of students who take none of the languages is 100-68 = 32.

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Answer:

Step-by-step explanation:

Set it up as

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4 years ago

What is 3 days 8 hours and 15 minutes plus 2 days 15 hours and 45 minutes?

kondaur [170]

A Total of 6 Days!

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