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Alex_Xolod [135]
3 years ago
5

Find the amount in an account if $7,000 is invested and is compounded continuously for 5 years at a 2% interest rate

Mathematics
1 answer:
Korvikt [17]3 years ago
7 0
<span>(See attached formula)
("e" = 2.718281828)
Total = principal * e ^ (rate * years)
</span><span>Total = 7,000 * e^(.02*5)
</span>Total = 7,000 * <span>2.718281828^.1
</span>Total = <span> <span> </span></span><span><span><span>7,000 * 1.1051709181 </span> </span> </span>
<span><span><span>Total = 7,736.20
 </span></span></span>

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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
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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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3 years ago
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Step-by-step explanation:

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1. You would divide the 33/3 first to get 11

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Your answer would be 8.

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Step-by-step explanation:

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