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2 years ago

Find the sum of the geometric series 40 + 40(1.005) + 40(1.005)^2 + ⋯ + 40(1.005)^11.

Mathematics

1 answer:

KiRa [710]2 years ago

4 0

Answer:

The sum is 493.4

Step-by-step explanation:

In order to find the value of the sum, you have to apply the geometric series formula, which is:

$\sum_{i=1}^{n} ar^{i-1} = \frac{a(1-r^{n})}{1-r}$

where i is the starting point, n is the number of terms, a is the first term and r is the common ratio.

The finite geometric series converges to the expression in the right side of the equation. Therefore, you don't need to calculate all the terms. You can use the expression directly.

In this case:

a=40

b= 1.005

n=12 (because the first term is 40 and the last term is 40(1.005)^11 )

Replacing in the formula:

$\frac{a(1-r^{n})}{1-r} = \frac{40(1-1.005^{12})}{1-1.005}$

Solving it:

The sum is 493.4

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Answer:

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And that represent the proportion of failures.

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Step-by-step explanation:

Previous concept

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.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

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Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by: