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2 years ago

How maany 0.5 ltres make 50 litres

Mathematics

2 answers:

AysviL [449]2 years ago

6 0

100
Simple:

0.5×100=50

elena-14-01-66 [18.8K]2 years ago

3 0

5 L169.0701135 fl oz to 50 L1690.701135 fl oz

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Step-by-step explanation:

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The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day.

valina [46]

Answer:

A sample of 179 is needed.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.85}{2} = 0.075$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.075 = 0.925$ , so Z = 1.44.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

A previous study found that for an average family the variance is 1.69 gallon?

This means that $\sigma = \sqrt{1.69} = 1.3$

If they are using a 85% level of confidence, how large of a sample is required to estimate the mean usage of water?

A sample of n is needed, and n is found for M = 0.14. So

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$0.14 = 1.44\frac{1.3}{\sqrt{n}}$

$0.14\sqrt{n} = 1.44*1.3$

$\sqrt{n} = \frac{1.44*1.3}{0.14}$

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2 years ago

Assume that all interest is simple interestBerger Car rental borrow $8500 at 4% interest to cover the increase cost of the auto

viva [34]

we know that

The formula of simple interest is equal to

$I=P(rt)$

In this problem

we have

I=$170

P=$8,500

r=4%=4/100=0.04

t=?

substitute given values in the formula

$\begin{gathered} 170=8,500(0.04t) \\ Solve\text{ for t} \\ t=\frac{170}{8,500*0.04} \\ t=0.5\text{ years} \end{gathered}$

therefore

0.5 years=6 months

<h2>The answer is 6 months</h2>

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10 months ago

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choli [55]

Answer:

By excluding (10, 15), a better description can be given for the data set.

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Overall, this data seems to suggest a decreasing trend. The value (10, 15) is an outlier; if we remove it, the trend appears to be stronger. Thus if we exclude this value we can give a better description of the data set.

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