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3 years ago

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A trapezoid has a height of 3 meters, a base length of 8 meters, and an area of 30 square meters. What is the length of the base

?

1 answer:

algol [13]3 years ago

4 0

Answer:

Length of the second base will be 12 meters.

Step-by-step explanation:

Area of a trapezoid is given by the formula,

Area = $\frac{1}{2}(b_1+b_2)h$

Here, b₁ and b₂ are the bases of the trapezoid and 'h' is the vertical distance between the bases.

By substituting the values in the formula,

$30=\frac{1}{2}(8+b_2)3$

$60=3(8+b_2)$

$20=8+b_2$

$b_2=12$

Therefore, length of the second base will be 12 meters.

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Wind farm configuration is a significant issue as the upwind turbines generate maximal energy while creating wakes for downwind

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Answer:

a) [24.114,29.2858]

b) Since 30 is not in the 95% confidence interval, there is a 95% probability that 30 is not the true mean WEP

Step-by-step explanation:

a)

The 95% confidence interval is given by the interval

$\bf [ \bar x-z^*\frac{s}{\sqrt n}, \bar x+t^*\frac{s}{\sqrt n}]$

where

$\bf \bar x$ = 26.7 is the sample mean

s = 17.7 is the sample standard deviation

n = 180 is the sample size

Since the sample size is big enough, we can use the Normal N(0,1) to compute $\bf z^*$ and it would be 1.96(*) (a value such that the area under the Normal curve outside the interval [-z, z] is 5% (0.05))

and our 95% confidence interval is

$\bf [26.7-1.96*\frac{17.7}{\sqrt{180}}, 26.7+1.96*\frac{17.7}{\sqrt{180}}]=\boxed{[24.114,29.2858]}$

(*)

This value can be computed in Excel with

<em>NORMINV(1-0.025,0,1)</em>

and in OpenOffice Calc with

<em>NORMINV(1-0.025;0;1)</em>

b)

Since 30 is not in the 95% confidence interval, there is a 95% probability that 30 is not the true mean WEP

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3 years ago

What are the comon factors of 70&20??

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4 years ago

Is education related to programming preference when watching TV? From a poll of 80 television viewers, the following data have b

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Answer:

a) H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

b) $\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

c) $\chi^2_{crit}=5.991$

d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

High school Some College Bachelor or higher Total

Public Broadcasting 15 15 10 40

Commercial stations 5 25 10 40

Total 20 40 20 80

We need to conduct a chi square test in order to check the following hypothesis:

Part a

H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

The level os significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part b

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{20*40}{80}=10$

$E_{2} =\frac{40*40}{80}=20$

$E_{3} =\frac{20*40}{80}=10$

$E_{4} =\frac{20*40}{80}=10$

$E_{5} =\frac{40*40}{80}=20$

$E_{6} =\frac{20*40}{80}=10$

And the expected values are given by:

High school Some College Bachelor or higher Total

Public Broadcasting 10 20 10 40

Commercial stations 10 10 20 40

Total 20 30 30 80

Part b

And now we can calculate the statistic:

$\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(2-1)(3-1)=2$

Part c

In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is $\chi^2_{crit}=5.991$

Part d

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >33.75)=2.23x10^{-7}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(33.75,2,TRUE)"

Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

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4 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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