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VladimirAG [237]

2 years ago

6

A triangle has a side length of 4in in an area of 18 square inches and a larger similar triangle has a corresponding side length

of 8 in . Determine the area of the larger triangle

1 answer:

Harman [31]2 years ago

5 0

Answer:

72

<em>s1</em><em> </em><em>:</em><em>s2</em><em> </em><em>=</em><em> </em><em>A1</em><em> </em><em>:</em><em> </em><em>A2^</em><em>2</em>

<em>So</em><em> </em><em>the</em><em> </em><em>factor</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>sides</em><em> </em><em>is</em><em> </em><em>2</em><em>.</em><em> </em><em>Now</em><em> </em><em>you</em><em> </em><em>multiply</em><em> </em><em>the</em><em> </em><em>area</em><em> </em><em>by</em><em> </em><em>2</em><em> </em><em>squared</em><em>.</em>

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Rainbow [258]

Answer: C. $(29,\ 37.8)$

Step-by-step explanation:

The confidence interval for difference of two population mean is given by :-

$\overline{x}_1-\overline{x}_2\pm z_{\alpha/2}\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}$

Given : Level of significance : $1-\alpha:0.99$

Then , significance level : $\alpha: 1-0.99=0.01$

Critical value : $z_{\alpha/2}=z_{0.005}=2.576$

$n_1=30\ ;\ n_2=50\\\\\overline{x}_1=46.7\ ;\ \overline{x}_2=13.3\\\\s_1=9.2\ ;\ s_2=1.9$

$46.7-13.3\pm(2.576)\sqrt{\dfrac{9.2^2}{30}+\dfrac{1.9^2}{50}}\approx33.4\pm4.38=(29.02\ ,37.78)\approx(29,\ 37.8)$

Hence, a 99% confidence interval for the difference between the mean wait times of everyone who rode both rides $(29,\ 37.8)$

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Step-by-step explanation:

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guapka [62]

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