Find each angle measure

Mathematics

1 answer:

Aloiza [94]3 years ago

4 0

It would surely be helpful if you let us know which angles to measure!

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BRAINLIEST ASAP! PLEASE ANSWER!!!!!

kenny6666 [7]

Similarities: They both are polynomials of degree 2, both of their graphs is a parabola, both have either 2 or 0 real solutions, they are both continuous functions over R

<span>(DOS= difference of two squares, PST=perfect square trinomial </span>

<span>Differences: PST has three terms, whereas the difference of squares has 2. PST's factors are both the same, whereas DOS's elements are conjugates of each other. DOS can always be factored into two distinct polynomials with rational coefficients, whereas PST has two same polynomial factors.</span>

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3 years ago

A random sample of 200 teenagers participated in a taste test. Each teenager sampled four choices of fruit drink (labeled A, B,

Vedmedyk [2.9K]

Answer:

it is c hope it helps have a good day

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3 years ago

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.