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4 years ago

Is a parallelogram a trapezoid always or not always?

Mathematics

1 answer:

Nina [5.8K]4 years ago

4 0

Answer:

No, a trapezoid is not a parallelogram. A trapezoid is defined as having Exactly two parallel sides while a parallelogram has two pairs of parallel sides.

Step-by-step explanation:

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Use the normal distribution to find a confidence interval for a difference in proportions p1-p2 given the relevant sample result

9966 [12]

Answer:

a) $\hat p_1 -\hat p_2= 0.2-0.35= -0.15$

b) $ME= 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.169$

c) $(0.2-0.35) - 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =-0.319$

$(0.2-0.35) + 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.0185$

And the 99% confidence interval would be given (-0.319;0.0185).

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".

$p_1$ represent the real population proportion 1

$\hat p_1=0.2$ represent the estimated proportion 1

$n_1=60$ is the sample size required 1

$p_2$ represent the real population proportion for 2

$\hat p_2 =0.35$ represent the estimated proportion 2

$n_2=100$ is the sample size required for Brand B

$z$ represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution

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

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_2)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

Part a

The best estimate is given by:

$\hat p_1 -\hat p_2= 0.2-0.35= -0.15$

Part b

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.

$z_{\alpha/2}=2.58$

The margin of error is given by:

$ME= 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.169$

Part c

And replacing into the confidence interval formula we got:

$(0.2-0.35) - 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =-0.319$

$(0.2-0.35) + 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.0185$

And the 99% confidence interval would be given (-0.319;0.0185).

4 0

4 years ago

kelli and her faimly went to the beach for a vacation. they drove 293 miles 7 hours to get theredid they drive each hour

Inessa [10]

Ok so, Since Kelli and her family drove a total of 293 miles in 7 hours, then your equation should be $\frac{293miles}{7hrs}$
Now, since you have your ratio you divide 293 by 7 in order to get how many miles were driven in one hour
$\frac{293}{7}$ = 41.85714286 miles/hr

5 0

3 years ago

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Analyze the diagram below and answer the question that follow

andrew-mc [135]

Answer:

Step-by-step explanation:

3 0

3 years ago

Find the midpoint M of the line segment joining the points P= (-7,-8) and Q = (3,2)

Fed [463]

Answer: