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2 years ago

7

The height of a trapezoid is 6 in. And its area is 48 in2. One base of the trapezoid is 6 inches longer than the other base. Wha

t are the lengths of the bases? Complete the explanation of how you found your answer.

1 answer:

omeli [17]2 years ago

4 0

Answer:

5 in
11 in

Step-by-step explanation:

Let x represent the length of the shorter base in inches. Then the longer base has length x+6. The area of the trapezoid is given by the formula ...

A = (1/2)(b1 +b2)h

Filling in the values we know, we have ...

48 = (1/2)(x +(x+6))(6)

16 = 2x +6 . . . . . divide by 3

10 = 2x . . . . . . . . subtract 6

5 = x . . . . . . . . . . divide by 2

(x+6) = 11 . . . . . . find the longer base

The lengths of the bases are 5 inches and 11 inches. We found them by solving an equation relating area to base length.

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Complete Question

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a

$SE = 0.66}$

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Step-by-step explanation:

From the question we are told that

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The first sample mean is $\= x _1 = 8$

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The first variance is $v_1 = 0.25$

The first variance is $v_2 = 0.55$

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Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the first standard deviation is

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=> $\sigma_1 = \sqrt{0.25}$

=> $\sigma_1 = 0.5$

Generally the second standard deviation is

$\sigma_2 = \sqrt{v_2}$

=> $\sigma_2 = \sqrt{0.55}$

=> $\sigma_2 = 0.742$

Generally the first standard error is

$SE_1 = \frac{\sigma_1}{\sqrt{n} }$

$SE_1 = \frac{0.5}{\sqrt{60} }$

$SE_1 = 0.06$

Generally the second standard error is

$SE_2 = \frac{\sigma_2}{\sqrt{n} }$

$SE_2 = \frac{0.742}{\sqrt{60} }$

$SE_2 = 0.09$

Generally the standard error of the difference between their mean scores is mathematically represented as

$SE = \sqrt{SE_1^2 + SE_2^2 }$

=> $SE = \sqrt{0.06^2 +0.09^2 }$

=> $SE = 0.66}$

Generally 95% confidence interval is mathematically represented as

$(\= x_1 -\= x_2) -(Z_{\frac{\alpha }{2} } * SE) < \mu_1 - \mu_2 < (\= x_1 -\= x_2) +(Z_{\frac{\alpha }{2} } * SE)$

=> $(8 -10) -(1.96 * 0.66) < \mu_1 - \mu_2 < (8-10) +(Z_{\frac{\alpha }{2} } * 0.66)$

=> $-3.29 < \mu_1 - \mu_2 < -0.70$

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