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Afina-wow [57]
3 years ago
7

Dan invests £1200 into his bank account. He receives 5% per year compound interest. How much will dan have after5 years?

Mathematics
1 answer:
solmaris [256]3 years ago
7 0

Answer:

Dan will have $1,531.53 after 5 years.

Step-by-step explanation:

To find the answer, you can use the following formula to calculate the future value:

F= P(1 + r)^t

F= Future value

P= Present value= 1200

r= rate of interest= 5%

t= time= 5

F=1200(1+0.05)^5

A=1200(1.05)^5

A=1531.53

According to this, the answer is that Dan will have $1,531.53 after 5 years.

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Answer:

4.4

Step-by-step explanation:

The parent function of this graph is:  y = sin(x)

The sine function is periodic, meaning it repeats forever.

Standard form of a sine function:

\sf f(x)= A \sin(B(x+C))+D

where:

  • A = amplitude (height from the mid-line to the peak)
  • 2π/B = period (horizontal distance between consecutive peaks)
  • C = phase shift (horizontal shift - positive is to the left)
  • D = vertical shift

The <u>period</u> is the horizontal distance between consecutive peaks, which is the same as <u>twice the horizontal distance between the intersection of the curve and the mid-line</u>.

Given consecutive points of intersection between the curve and the mid-line:

  • (3.7, 5)
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Therefore, the horizontal distance between these two points is:

5.9 - 3.7 = 2.2

⇒ Period = 2.2 × 2 = 4.4

-------------------------------------------------------------------------------------

To create the equation for the function.

From inspection of the given graph:

  • Amplitude (A) = 6
  • Mid-line is y = 5
  • Vertical shift (D) = +5

Period = 2π/B = 4.4  ⇒  B = 5π/11

Phase Shift (C) = -3.7

Substituting the values into the standard form:

\implies \sf f(x)= 6 \sin\left(\dfrac{5}{11}\pi(x-3.7)\right)+5

\implies \sf f(x)= 6 \sin\left(\dfrac{5}{11} \pi x-\dfrac{37}{22}\pi \right)+5

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Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th
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(a) 100 fishes

(b) t = 10: 483 fishes

    t = 20: 999 fishes

    t = 30: 1168 fishes

(c)

P(\infty) = 1200

Step-by-step explanation:

Given

P(t) =\frac{d}{1+ke^-{ct}}

d = 1200\\k = 11\\c=0.2

Solving (a): Fishes at t = 0

This gives:

P(0) =\frac{1200}{1+11*e^-{0.2*0}}

P(0) =\frac{1200}{1+11*e^-{0}}

P(0) =\frac{1200}{1+11*1}

P(0) =\frac{1200}{1+11}

P(0) =\frac{1200}{12}

P(0) = 100

Solving (a): Fishes at t = 10, 20, 30

t = 10

P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483

t = 20

P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999

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P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168

Solving (c): \lim_{t \to \infty} P(t)

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of e^{-ct} decreases

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{t \to \infty} = {e^{-ct} \to 0}

So:

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P(\infty) = \frac{1200}{1 + 11 * 0}

P(\infty) = \frac{1200}{1 + 0}

P(\infty) = \frac{1200}{1}

P(\infty) = 1200

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