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3 years ago

Easy maths question I’ll give brainiest

Mathematics

2 answers:

NISA [10]3 years ago

5 0

Answer:

BC ≈ 12.6 m

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan31° = $\frac{opposite}{adjacent}$ = $\frac{AB}{BC}$ = $\frac{7.6}{BC}$ ( multiply both sides by BC )

BC × tan31° = 7.6 ( divide both sides by tan31° )

BC = $\frac{7.6}{tan31}$ ≈ 12.6 m ( to 3 sf )

LUCKY_DIMON [66]3 years ago

4 0

Answer:

Angle BC is 12.6m

Step-by-step explanation:

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3 years ago

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 - <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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3 years ago