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leonid [27]
3 years ago
7

If $4,000 is deposited in a bank account paying 4% compounded quarterly, what amount will be in the account after 7 years? How m

uch interest will be returned during the 7 years?
Mathematics
1 answer:
pishuonlain [190]3 years ago
8 0
We shall use the equation A = i(1 + r/n)^nt
So we shall plug all of that in so it turns out like this:
A = 4000(1 + 0.04/4)^4 x 7

So then that can be simplified to A = (1.01)^28
and then to A = 4000(1.321)
so then A = 5285.16
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R^3 +4r^2-25r-100<br><br> Help
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Answer:

The zeroes in this equation are -5, -4, and 5

Step-by-step explanation:

In order to find these, you need to factor by splitting. For this, we separate out the two halves of the equation and pull out the greatest common factor of each. Let's start with the front end.

r^3 + 4r^2

r^2(r + 4)

Now the second half.

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-25(r + 4)

Since what is left in the parenthesis are exactly the same, we can use that parenthesis next to one with what we pulled out.

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Now that we are fully factored, set each parenthesis equal to 0 and solve for x.

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3 years ago
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A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears
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Answer:

(a) The probability that <em>Y</em>₂₀ exceeds 1000  is 3.91 × 10⁻⁶.

(b) <em>n</em> = 28.09

Step-by-step explanation:

The random variable <em>Y</em>ₙ is defined as the total numbers of dollars paid in <em>n</em> years.

It is provided that <em>Y</em>ₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of <em>Y</em>ₙ are:

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(a)

For <em>n</em> = 20 the mean and standard deviation of <em>Y</em>₂₀ are:

\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\

Compute the probability that <em>Y</em>₂₀ exceeds 1000 as follows:

P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z>  4.47)\\=1-P(Z

**Use a <em>z </em>table for probability.

Thus, the probability that <em>Y</em>₂₀ exceeds 1000  is 3.91 × 10⁻⁶.

(b)

It is provided that P (<em>Y</em>ₙ > 1000) > 0.99.

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Compute the value of <em>n</em> as follows:

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The roots of a quadratic equation are:

n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of <em>n</em> = 28.09.

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Answer:

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