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PLEASEEEEE HELP…
In Exploration 3.1.2 you went through a procedure that helped you see how a formula
with n compounds per year relates to the formula for compounding continuously. Select
the appropriate answers to model what you did in the exploration.
Annual growth Rate
A (t) = P(1 + r)
n compounds per year
nt
A () = (1+ a)
a =
[Select]
n
let z = -
7
A (t) = P(1 + [b])"
b = (Select)
let n = ar
Ict
C =
[Select)
A (0) P(1)
110) - P{(x - 2)
+
2

PLEASEEEEE HELP In Exploration 312 You Went Through A Procedure That Helped You See How A Formula With N Compounds Per Year Relates To The Formula For Compoundi class=

Sagot :

The growth amount is given by the growth rate the duration of the growth

and how the rate is applied to each period.

The correct selections are;

  • a = n
  • [tex]\displaystyle b = \frac{1}{x}[/tex]
  • c = x·r
  • [tex]\displaystyle \lim_{x \to \infty} \left(1 + \frac{1}{x} \right)^x \to 1[/tex]

Reasons:

First part;

The given formula for the annual growth rate is; [tex]A(t) = \mathbf{P \cdot \left(1 + r)^t}[/tex]

[tex]\displaystyle A(t) = \mathbf{P \cdot \left (1 + \frac{r}{[a]} \right)^{n \cdot t}}[/tex]

Where;

P = The principal

r = The rate per period

n = The number of compounding per period

t = The number of periods

In the above formula, we have that the number of compounding per periods = n

Therefore;

  • a = n

Which gives the fraction of the interest applied to each period as [tex]\displaystyle \frac{r}{n}[/tex]

Which gives;

[tex]\displaystyle A(t) = \mathbf{P \cdot \left (1 + \frac{r}{n} \right)^{n \cdot t}}[/tex]

Second part;

When [tex]x = \displaystyle \frac{n}{r}[/tex], we have; [tex]A(t) = \mathbf{P \cdot \left(1 + [b])^{n \cdot t}}[/tex]

[tex]\displaystyle A(t) = P \cdot \left (1 + \frac{r}{n} \right)^{n \cdot t}[/tex]

Therefore;

[tex]\displaystyle b=\frac{r}{n}[/tex]

[tex]\displaystyle \frac{1}{x} = \frac{1}{\left(\frac{n}{r} \right)} =\frac{r}{n}[/tex]

Which gives;

[tex]\displaystyle b=\frac{r}{n} = \mathbf{\frac{1}{x}}[/tex]

[tex]\displaystyle \underline{ b = \frac{1}{x} }[/tex]

Third part;

Where, n = x·r, and [tex]\displaystyle A(t) = \mathbf{P \cdot \left (1 + \frac{1}{x} \right)^{[c] \cdot t}}[/tex]

We have;

[tex]\displaystyle x= \frac{n}{r}[/tex]

[tex]\displaystyle \frac{1}{x}=\frac{r}{n}[/tex]

[tex]\displaystyle A(t) = P \cdot \left (1 + \frac{1}{x} \right)^{[c] \cdot t} = \mathbf{P \cdot \left (1 + \frac{r}{n} \right)^{[c] \cdot t}}[/tex]

Which gives;

c = n = x·r

  • c = x·r

Fourth part;

If x is a really large number, we have; [tex]\displaystyle \mathbf{\left(1 + \frac{1}{x} \right)^x}[/tex]

Where x approaches ∞, we have;

[tex]\displaystyle \frac{1}{\infty}=0[/tex]

Which gives;

[tex]\displaystyle \displaystyle \left(1 + \frac{1}{x} \right)^x= \left(1 + \frac{1}{\infty} \right)^\infty = \displaystyle \left(1 + 0\right)^\infty = 1 ^\infty = 1[/tex]

  • [tex]\displaystyle \underline{\lim_{x \to \infty} \left(1 + \frac{1}{x} \right)^x \to 1}[/tex]

Learn more about growth rate formula here:

https://brainly.com/question/2284816

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