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TiliK225 [7]
4 years ago
7

A student loan you took out is due in 3 years and requires repayment of $3,000. To pay off the loan, you set aside money in an a

ccount that pays 7% interest.
To meet your obligation, how much money do you need to deposit in the account?
Mathematics
1 answer:
damaskus [11]4 years ago
7 0

Answer:

P = $2448.89

P ~= $2,449

He need to deposit $2,449

Step-by-step explanation:

Given:

Interest rate r= 7% = 0.07

Number of years n = 3 years

Future value that should be meet A = $3000

We need to calculate the initial investment (Principal P). Using the compound interest formula:

A = P(1+r)^n

P = A/(1+r)^n

Substituting the values of A, r, n, we have;

P = 3000/(1+0.07)^3

P = $2448.89

P ~= $2,449

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-7 is the slope
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A wheel had a radius of 15cm . Approximately how far does it travel in 4 revolutions
m_a_m_a [10]

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Step-by-step explanation:

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3 years ago
A bag contains 11 numbered chips inside. Two chips are labeled 1, three are labelled 2, and six are labelled 3. Let the random v
Dima020 [189]

Answer: 2.36

Step-by-step explanation:

Using the μ=∑[x⋅P(X=x)

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6/11 because you have 6 labeled 3

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Hope that helps, plz put a good rating

4 0
3 years ago
1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves
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(a) The differential equation is separable, so we separate the variables and integrate:

(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx

\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx

\ln|y| = x - \ln|x+1| + C

When x = 0, we have y = 2, so we solve for the constant C :

\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)

Then the particular solution to the DE is

\ln|y| = x - \ln|x+1| + \ln(2)

We can go on to solve explicitly for y in terms of x :

e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}

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x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1

and the bounded region is the set

\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}

The area of this region is

\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}

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3 years ago
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MP,
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