Lin ran 2 3/4 miles in 2/5 of an hour, Noah ran 8 2/3 miles in 4/3 of an hour, how far would Lin run in 1 hour?

Mathematics

1 answer:

Marizza181 [45]2 years ago

5 0

Answer:

$6\frac{7}{8}\ miles$

Step-by-step explanation:

Given that Lil's distance in 2/5 hrs is 2 3/4hrs

-Let x be the distance ran in 1 hr.

#We equate and cross multiply to solve for x as follows:

$\frac{2}{5}\ hrs=2\frac{3}{4}\\1\ hr=x\\\\\therefore x=2\frac{3}{4}/\frac{2}{5}\\\\=6\frac{7}{8}\ miles$

Hence, Lil can run 6 7/8 miles in 1 hr.

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

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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 - α)% 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:

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

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

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

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

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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