Find the domain and range of the relation and determine whether it is a function

Mathematics

2 answers:

Ne4ueva [31]3 years ago

5 0

The domain is going to be all ur x values...so the domain is { -4,-1,1,4 }....the range will be all ur y values...so the range is { -1,1,4} ...the reason there is only 3 numbers is because if the numbers repeat, u only have to write them once.

A function will not have any repeating x values....they can have repeating y values, just not the x ones. So if all ur x values are different, then it is a function. Therefore, this is a function.

fgiga [73]3 years ago

4 0

Answer:

A. Domain -4,-1,4,1 ; Range -1,1,4 ; Yes, the relation is a function.

Step-by-step explanation:

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There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

Ann [662]

Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

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

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

The confidence interval would be given by this formula

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

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.