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andriy [413]
3 years ago
8

Rita is saving up $15,000 to put a down payment on a condominium. if she starts with $8000 saved and saves an additional $850 ea

ch month, which equation represents how far rita is from her goal of reaching $15,000 saved? let x stand for months and y stand for dollars.
Mathematics
1 answer:
gavmur [86]3 years ago
8 0
If she starts with $8000, she has only $7000 to go.This is reduced by $850 per month.
$7000 - $850x
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Please help me with the below question.
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By letting

y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}

we get derivatives

y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}

y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0

Examine the lowest degree term \left(x^{r-1}\right), which gives rise to the indicial equation,

5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients c_k is

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so that with r = 4/5, the coefficients are governed by

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c) Starting with c_0=1, we find

c_1 = -\dfrac{c_0}5 = -\dfrac15

c_2 = -\dfrac{c_1}{10} = \dfrac1{50}

so that the first three terms of the solution are

\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}

4 0
2 years ago
assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. find the
drek231 [11]

Answer:

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Step-by-step explanation:

Let X be the height of a woman randomly choosen. We know tha X have a mean of 63.6 inches and a standard deviation of 2.5 inches. For an x value, the related z-score is given by z = (x-63.6)/2.5. We are looking for a value x_{0} such that P(X < x_{0}) = 0.76, but, 0.76 = P(X < x_{0}) = P((X-63.6)/2.5 < (x_{0}-63.6)/2.5) = P(Z < (x_{0}-63.6)/2.5), i.e., (x_{0}-63.6)/2.5 is the 76th percentile of the standard normal distribution. So, (x_{0}-63.6)/2.5 = 0.7063, x_{0} =63.6+(2.5)(0.7063) = 65.3658. Therefore, the height of a woman who is at the 76th percentile is 65.3658 inches.

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saw5 [17]

Answer:

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