If you save the same amount each week, the amount of money

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

4 0

Answer:

I don't really get your question. But if you save the same amount each week. The amount of money will be the amount you save each week multiplied by the number of weeks

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The number obtained on rationalizing the denominator of 1/(√7--2) is

Slav-nsk [51]

$\frac{1}{ \sqrt{7} -(- 2)} \\ rationalizing \: \: \: the \: \: \: denominator \: \: \: we \: \: \: have \\ = \frac{1}{ \sqrt{7} + 2 } \times \frac{ \sqrt{7} - 2}{ \sqrt{7} - 2} \\ = \frac{ \sqrt{7} - 2}{ {( \sqrt{7} )}^{2} - {(2)}^{2} } \\ = \frac{ \sqrt{7} - 2 }{7 - 4} \\ = \frac{ \sqrt{7} - 2}{3}$

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Doubt clarification - use comment section.

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2 years ago

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A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq

Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

$s=16.6$ represent the sample deviation

$\bar X=12.8$ represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

$\alpha=1-0.99=0.01$ represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

$\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.