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kotegsom [21]
4 years ago
12

Doris put $4000 in a 5­year CD paying 6% interest compounded monthly. After 2 years, she withdrew all her money, and as an early

withdrawal penalty, she paid back all the interest she made during the first year. How much money was Doris left with? Show your work.
Mathematics
2 answers:
aleksandr82 [10.1K]4 years ago
8 0

Answer:  Doris was left with $4216.93.

Step-by-step explanation:

Since we have given that

Principal amount = $ 4000

Rate of interest = 6%

He put $4000 in a 5 year CD paying 6% interest compounded monthly.

After 2 years, she withdrew all her money.

so, Amount after 2 years when it is compounded annually.

A=4000(1+\dfrac{6}{12\times 100})^{12\times 2}\\\\A=4000(1+0.005)^{24}\\\\A=4000(1.005)^{24}\\\\A=\$4508.64

As an early withdrawal penalty, she paid back all the interest she made during the first year.

Interest=4000[(1+0.005)^{12}-1]\\\\Interest=\$246.71

Amount left with Doris is given by

\$4508.64-\$246.71\\\\=\$4261.93

Hence, Doris was left with $4216.93.

sp2606 [1]4 years ago
5 0
For APEX it is going to be 4261.93

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A vet borrows $12,000 for 4 years. The total interest paid was $960. What was the interest rate on the loan?
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Answer:

2% per year.

Step-by-step explanation:

Solving our equation

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3 years ago
A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement
dsp73

Answer:

P(First\ White\ and\ Second\ Blue) = \frac{3}{28}

P(Same) = \frac{67}{210}

Step-by-step explanation:

Given (Omitted from the question)

Red = 7

White = 9

Blue = 5

Solving (a): P(First\ White\ and\ Second\ Blue)

This is calculated using:

P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)

P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}

<em>We used Total - 1 because it is a probability without replacement</em>

So, we have:

P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}

P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}

P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}

P(First\ White\ and\ Second\ Blue) = \frac{45}{420}

P(First\ White\ and\ Second\ Blue) = \frac{3}{28}

Solving (b) P(Same)

This is calculated as:

P(Same) = P(First\ Blue\ and Second\ Blue)\or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)

P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})

P(Same) = (\frac{5}{21} * \frac{4}{20})+(\frac{7}{21} * \frac{6}{20})+(\frac{9}{21} * \frac{8}{20})

P(Same) = \frac{20}{420}+\frac{42}{420} +\frac{72}{420}

P(Same) = \frac{20+42+72}{420}

P(Same) = \frac{134}{420}

P(Same) = \frac{67}{210}

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Step-by-step explanation:

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