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3 years ago

Subtract. 1/6 −(−7 1/3 )

Mathematics

1 answer:

Leokris [45]3 years ago

4 0

Answer:

The answer is 7 1/2 or 15/2 or 7.5 (all the same just in mixed fraction and decimal form)

Step-by-step explanation:

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Step-by-step explanation:

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2 years ago

-2x-4+5x=8 please help, I'm too stupid to do this.

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Answer: x=1 1/3

Step-by-step explanation:

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3 years ago

Read 2 more answers

Your state is considering raising the legal age for consumption of alcoholic beverages to 21 years old. How large a sample size

julsineya [31]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.01}{2.58})^2}=16641$

And rounded up we have that n=16641

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

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:

$z_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

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 $ME =\pm 0.01$ 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)

For this case we can use as estimator for the proportion $\hat p =0.5$ , since we don't have any other previous info. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.01}{2.58})^2}=16641$

And rounded up we have that n=16641

8 0

3 years ago

The American Management Association is studying the income of store managers in the retail industry. A random sample of 49 manag

VashaNatasha [74]

Answer:

a) The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

b) The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

Step-by-step explanation:

Question a:

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

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a p-value of $1 - 0.025 = 0.975$ , so Z = 1.96.

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.

$M = 1.96\frac{2050}{\sqrt{49}} = 574$

The lower end of the interval is the sample mean subtracted by M. So it is 45420 - 574 = $44,846.

The upper end of the interval is the sample mean added to M. So it is 45420 + 574 = $45,994.

The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

Question b:

The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

5 0

3 years ago