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3 years ago

Which Doesn't Belong?

Mathematics

2 answers:

klio [65]3 years ago

8 0

I’m assuming 9? There isn’t a question so

Kryger [21]3 years ago

8 0

Step-by-step explanation:

If you are goin for which one isn't prime, then it would be four.

If you are going for which one isnt a factor, there is more then one

If you are going for something else, i have no idea.

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Graph the hyperbola y^2/16-x^2/9=1. What type of transverse axis does it have?

gulaghasi [49]

<span>Center(0,0)
a^2=16
a=4
b^2=9

b=3
c^2=a^2+b^2=16+9=25
c=√25=5</span>

3 0

3 years ago

Two identical rubber balls are dropped from different heights. Ball 1 is dropped from a height of 196 feet, and ball 2 is droppe

Anon25 [30]

H1 (t) = 196 - 16 t-squared. / / / H2 (t) = 271-16t-squared. / / / In each function, 't' is the number of seconds after that ball is dropped. / / / Each function is only true until the first time that H=0, that is, until the first bounce. Each function becomes very complicated after that, and we would need more information in order to write it.

5 0

3 years ago

Please help, thank you

Trava [24]

A=h•b/2
h=2x+1
b=2x
A=(2x+1) •2x/2
Divide numerator and denominator by 2
A=x(2x+1)

5 0

2 years ago

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: