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3 years ago

how many cubic sugar of edge 20cm can be fitted into cuboid container of dimension 40cm by 60cm by100cm

Mathematics

1 answer:

Sloan [31]3 years ago

8 0

Answer:

n =30

Step-by-step explanation:

Given that,

Size of a sugar cube = 20 cm

The dimension of the container is 40cm by 60cm by100cm.

Let n be the number of sugar cube that are contained in the container.

So,

$n=\dfrac{\text{volume of container}}{\text{volume of sugar cube}}\\\\n=\dfrac{40\times 60\times 100}{20^3}\\\\n=30$

So, there are 30 sugar cube that are fitted in the container.

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o-na [289]

Answer:

a) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401$

And rounded up we have that n=2401

b) $n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22$

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

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.

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".

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 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

Part a

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.02$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Since we don't have prior information about the population proportion we can use ase estimator $\hat p =0.5$ . And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401$

And rounded up we have that n=2401

Part b

For thi case the estimator for p is $\hat p =0.07$

$n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22$

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

4 0

3 years ago

how many cubic sugar of edge 20cm can be fitted into cuboid container of dimension 40cm by 60cm by100cm​

how many cubic sugar of edge 20cm can be fitted into cuboid container of dimension 40cm by 60cm by100cm