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valentinak56 [21]
3 years ago
13

Mara found the length of time of an investment. The principal of the investment was $4,300, the interest rate was 6.2 percent, a

nd the interest was $2,666. Mara made an error in her work.
I = p r t. 2666 = (4300) (0.062) t. 2666 = (266.6) t. StartFraction 266.6 over 2666 EndFraction = t. 0.1 = t.
Mathematics
1 answer:
77julia77 [94]3 years ago
3 0

well, 6.2% of 4,300 can't be over half. she has made a calculation error, and the real answer is (this is where the mistake was made) 4,300 * 0.062 (not 6.2, because percents are always divided by 100 to make them into a usable number), which is equal to $266.60.

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Solve these compound interest problems and round your answer to the nearest 100th.
MatroZZZ [7]

Answer:

$2,226.96

Step-by-step explanation:

You are going to want to use the compound interest formula, which is shown below.

A=P(1+\frac{r}{n} )^{nt}

<em>P = initial balance </em>

<em>r = interest rate </em>

<em>n = number of times compounded annually </em>

<em>t = time </em>

<em />

First, change 10% into its decimal form:

10% -> \frac{10}{100} -> 0.1

Now lets plug in the values into the equation:

A=500(1+\frac{0.1}{12})^{15(12)}

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The final amount after 15 years is $2,226.96

4 0
3 years ago
on a large college campus, 84% of the students report drinking alcohol within the past month, 33% report using some type of toba
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Answer:  0.31

Step-by-step explanation:

Let A denotes the event that the students report drinking alcohol and B denotes the students report using some type of tobacco product .

Given : P(A) =0.84   ; P(B)=0.33    and        P(A∪B)=0.86

We know that P(A\cap B)=P(A)+P(B)-P(A\cup B)

Then, the probability that the student both drunk alcohol and used tobacco in the past month is given by :-

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Hence, the probability that the student both drunk alcohol and used tobacco in the past month = 0.31

5 0
2 years ago
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