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jeka57 [31]
3 years ago
9

amelia is going to invest in an account paying on interest rate of 6.5% compounded daily? help me plz someone

Mathematics
1 answer:
yaroslaw [1]3 years ago
7 0

Answer:

$9,763

Step-by-step explanation:

Using the compund interest formula;

A = P(1+r)^t

A is the amount = $10,200

rate r = 6.5%= 0.065

time t = years

n = 1/365 ((daily compounding)

Substitute into the formula and get Pincipal P

10,200 = P(1+0.065(365))^5/365

10200 = P(1+23.725)^0.01369

10200 = P(24.725)^0.01369

10200 = P(1.0448)

P = 10200/1.0448

P = 9,762.6

Hence the amount invested is $9,763 to nearest dollars

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What is the nth term rule of the quadratic sequence below?
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Answer:

3n² + 5n - 2

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<u>Given sequence</u>:

6, 20, 40, 66, 98, 136, ...

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6 \underset{+14}{\longrightarrow} 20 \underset{+20}{\longrightarrow} 40 \underset{+26}{\longrightarrow} 66 \underset{+32}{\longrightarrow} 98 \underset{+38}{\longrightarrow} 136

As the first differences are not the same, calculate the <u>second differences:</u>

14 \underset{+6}{\longrightarrow} 20 \underset{+6}{\longrightarrow} 26 \underset{+6}{\longrightarrow} 32 \underset{+6}{\longrightarrow} 38

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Compare 3n² with the given sequence:

\begin{array}{|c|c|c|c|c|}\cline{1-5} n & 1 & 2 & 3 & 4\\\cline{1-5} 3n^2 & 3 & 12 & 27 & 48 \\\cline{1-5} \sf operation & +3&+8 & +13 & +18 \\\cline{1-5} \sf sequence & 6 & 20 & 40 & 66\\\cline{1-5}\end{array}

The second operations are different, therefore calculate the differences <em>between</em> the second operations:

3 \underset{+5}{\longrightarrow} 8 \underset{+5}{\longrightarrow} 13\underset{+5}{\longrightarrow} 18

As the differences are the same, we need to add 5n as the second operation:

\begin{array}{|c|c|c|c|c|}\cline{1-5} n & 1 & 2 & 3 & 4\\\cline{1-5} 3n^2  +5n & 8&22 & 42 & 68\\\cline{1-5}\sf operation & -2 &-2  &-2  & -2  \\\cline{1-5} \sf sequence & 6 & 20 & 40 & 66\\\cline{1-5}\end{array}

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Therefore, the nth term of the quadratic sequence is:

3n² + 5n - 2

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