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enyata [817]
3 years ago
13

Which of the following is the interest on $6000 at 6% compounded semiannually for eight years?

Mathematics
1 answer:
ryzh [129]3 years ago
4 0

Answer:

$3628.24

Step-by-step explanation:

we use the formula for accrued value (A) with compounded interest:

A=P(1+\frac{r}{n})^{n*t}

where A= accrued value (principal plus the accumulated  interest)

P = principal -> in our case $6000

r = annual interest rate (in decimal form) -> in our case 0.06

n = number of compoundings per year. In our case 2 (semiannually)

t = time in years -> in our case 8

A=P(1+\frac{r}{n})^{n*t}\\A = 6000(1+\frac{0.06}{2})^{16}\\ A= 6000(1+0.03)^{16}\\A = $9628.24

Since this is the value of principal plus accumulated interest, we subtract from it the principal ($6000) to get the value of just the interest:

$9628.24 - $6000 = $3628.24

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a = g/c - r/c + d/c

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable

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The current population of a town is 10,000 and its growth in years can be represented by P(t) = 10,000(0.2)^t, where t is the ti
DiKsa [7]

Answer:

1) 20%

2) Choice a.

Step-by-step explanation:

P(t)=10000(0.2)^t

1) P(0) is the population initially.

P(1) is the population after a year.

\frac{P(1)}{P(0)} represents the population increase factor.

So let's evaluate that fraction:

\frac{P(1)}{P(0)}

\frac{10000(0.2)^1}{10000(0.2)^0}

\frac{0.2^1}{0.2^0}=\frac{0.2}{1}=0.2

0.2=20%

2) Let's figure out the population growth in terms of months instead of years.

P(t)=10000(0.2)^{t}

We want t to represent months.

A full year is 12 months, in a full year we have that P(1)=10000(0.2)^1=10000(0.2)=2000

So we want a new P such that P(12)=2000 since 12 months equals a year.

Let's look at the functions given to see which gives us this:

a) P(12)=10000(0.87449)^{12}=2000 \text{approximately}

b) P(12)=10000(0.87449)^{12(12)}=0 \text{ approximately}

c) P(12)=10000(0.87449)^{\frac{1}{12}}=9889 \text{approximately}

d) P(12)=10000(0.87449)^{12+12}=400 \text{approximately}

So a is the function we want.

Also another way to look at this:

P(t)=10000(.2)^t where t is in years.

P(t)=10000(.2^\frac{1}{12})^t where t is in months.

And .2^\frac{1}{12}=0.874485 \text{approximately}

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4 years ago
NEED HELP ASAP!!!!!!
UNO [17]

Answer:

I do not know. sorry I couldn't help

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