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igor_vitrenko [27]
3 years ago
8

$1800 is invested at 4% compounded annually. Find the future value of the investment in 5 years. Find the interest that the inve

stment will earn after the term.
Mathematics
2 answers:
harina [27]3 years ago
8 0
5 Years is 1800x1.04^5 = 2189.98
Interest is 389.98
Ivanshal [37]3 years ago
7 0

Answer: the interest that the investment will earn after the term is $38.97


Step-by-step explanation:

1,800 x .04 x 1/2 = 36<u> (interest earned after 1 year)</u>

1,800 + 36  = 1,836

1,836 x .04 x 1/2 = 36.72 <u>(interest earned after 2 years)</u>

1,836 + 36.72 = 1,872.72

1,872.72 x .04 x 1/2 = 37.4544 -> 37.45 <u>(interest earned after 3 years)</u>

1,872.72 + 37.45 = 1,910.17

1,910.17 x .04 x 1/2 = 38.2034 -> 38.20 <u>(interest earned after 4 years)</u>

1,910.17 + 38.20 = 1,948.37

1,948.37 x .04 x 1/2 = 38.9674 -> 38.97 <u>(interest earned after 5 years)</u>




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Answer:

The first option is not a direct variation

Step-by-step explanation:

When we talk of a direct variation, as one value increases, the other value increases too

Or as one value decreases, the other value decreases

A direct variation is of the form;

y = kx

k = y/x

where k is the coefficient of variation that must be a constant value all through the set of values

The values we are comparing here are the x and y values

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Answer:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

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Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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\bar X represent the sample mean

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

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And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

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