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3 years ago

Question and Answers below in photo!! Please answer! Will mark BRAINLIEST! ⬇⬇⬇⬇⬇⬇⬇

Mathematics

1 answer:

Aleks04 [339]3 years ago

5 0

Answer:

72 $yd^{3}$

Step-by-step explanation:

Volume for a triangular prism = $\frac{1}{2}$ a · c · h

A = Height. C = Width. H = Length.

A = 3 yd. C = 8 yd. H = 6 yd.

V = 114 yd.

V = 114 yd ÷ 2 yd = 72 $yd^{3}$ .

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AleksAgata [21]

Answer:

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3 years ago

Read 2 more answers

A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of <em>z</em> for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

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Answer:

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Step-by-step explanation:

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