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statuscvo [17]
3 years ago
6

Mia found the area of a polygon The area is 32 square cm.

Mathematics
1 answer:
solniwko [45]3 years ago
8 0

Answer:

Only the isosceles trapezoid has an area of 32 cm².

Step-by-step explanation:

Let's calculate the area of each polygon.

For the two triangles we have:

A_{t} = \frac{bh}{2} = \frac{2 cm*4 cm}{2} = 4 cm^{2}

This polygon does not have an area of 32 cm².

For the rectangle we have:

A_{r} = bh = 6 cm*4 cm = 24 cm^{2}

This polygon does not have an area of 32 cm².

For the rectangle trapezoid we have:

A_{rt} = A_{t} + A_{r} = (4 + 24) cm^{2} = 28 cm^{2}

So, this polygon does not have an area of 32 cm².

Finally, for the isosceles trapezoid:

A_{it} = A_{t}*2 + A_{r} = (4*2 + 24) cm^{2} = 32 cm^{2}

This polygon does have an area of 32 cm².

   

Therefore, only the isosceles trapezoid has an area of 32 cm².

I hope it helps you!                              

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Step-by-step explanation:

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3 years ago
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My favorite pizza restaurant advertises that their average delivery time is 20 minutes. It always feels like it takes forever fo
andrezito [222]

Answer:

Population parameter(s): mean μ=20

Sample statistics: M=20.7, s=2.1

Hypotheses:

H_0: \mu=20\\\\H_a:\mu> 20

Test Statistic: t=2.357

p-value: 0.011

Reject H? YES

Conclusion (in context of the problem): There is  enough evidence to support the claim that the advertised delivery time is overly optimistic and it is larger than 20 minutes.  

<em>Is this difference practically significant? </em>No, the difference as the sample mean delivery time is under one minute of difference from the advertised time. Although it is significantly from the statistical point of view, the  difference is under 5% of the advertised delivery time.

The 99% confidence interval for the mean is (19.904, 21.496).

With this level of confidence, the null hypothesis failed to be rejected, as t=20 is a possible value for the true mean (is included in the confidence interval).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the advertised delivery time is overly optimistic and it is larger than 20 minutes.  

Then, the null and alternative hypothesis are:

H_0: \mu=20\\\\H_a:\mu> 20

The significance level is 0.05.

The sample has a size n=50.

The sample mean is M=20.7.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=2.1.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{2.1}{\sqrt{50}}=0.297

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{20.7-20}{0.297}=\dfrac{0.7}{0.297}=2.357

The degrees of freedom for this sample size are:

df=n-1=50-1=49

This test is a right-tailed test, with 49 degrees of freedom and t=2.357, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>2.357)=0.011

As the P-value (0.011) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that the advertised delivery time is overly optimistic and it is larger than 20 minutes.  

2. We have to calculate a 99% confidence interval for the mean.

The sample mean is M=20.7.

The sample size is N=50.

The t-value for a 99% confidence interval is t=2.68.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.68 \cdot 0.297=0.796

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 20.7-0.796=19.904\\\\UL=M+t \cdot s_M = 20.7+0.796=21.496

The  99% confidence interval for the mean is (19.904, 21.496).

With this level of confidence, the null hypothesis failed to be rejected, as t=20 is a possible value for the true mean (is included in the confidence interval).

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The equation is:

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They are all on the same line.

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