PLEASE HELP 50 POINTS

Mathematics

2 answers:

Arada [10]3 years ago

8 0

Answer:

im leaning on parallelogram

Step-by-step explanation:

ZanzabumX [31]3 years ago

6 0

Parallelogram, Quadrilateral, and Rectangles

First off both of these images are parallelogram because the arrows show that both the rectangle and the square (diamond/whatever you want to call it) have pairs of parallel sides so we know that both these shapes are parallelograms

Secondly, these shapes are both quadrilaterals. A quadrilateral, by definition, is simply a four-sided figure. Both a rectangle and a square are four-sided figures so these are both quadrilaterals.

Lastly, these figures are both rectangles. Yes I know one is a square and one is a rectangle but a square is a special four sided figure that can classify as a rectangle. Every square is a rectangle but not every rectangle is a square. A rectangle by definition is a four sided polygon with two pairs of parallel sides, congruent sides, and four 90 degree interior angles.

Hope this helped :)

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The Ohio Department of Agriculture tested 203 fuel samples across the state

Rus_ich [418]

Answer:

$\hat p = \frac{14}{105}= 0.133$

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:

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

If we replace the values obtained we got:

$0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475$

$0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185$

The 99% confidence interval would be given by (0.0475;0.2185)

Step-by-step explanation:

Previous concept

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: