In a certain village of Nepal, all the people speak either Nepali, or Maithili

Mathematics

1 answer:

galina1969 [7]3 years ago

4 0

Given,

Nepali speaker : 80% (Y)

Maithili speaker : 60% (X)

We know that,

Both language speaker = Y -X

Now,

Y- X

= 80- 60

= 20

Therefore, 20% people speak both languages

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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 - 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:

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

The critical value of z 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).

5 0

3 years ago

85 x 63 whoever gets it gets braiest

Katena32 [7]

85 x 63 = 5355

hope it helps

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6 0

3 years ago

Read 2 more answers

How do I solve this proportion?? 7/n = 8 / 7

kirill115 [55]

$\frac{7}{n}=\frac{8}{7} \\\\ n=\frac{7*7}{8} \\\\ \boxed{n=\frac{49}{8}}$