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3 years ago

The expression sin 50 cos 40 + cos 50 sin 40 is equivalent to

1 answer:

8 0

Sin (a+b) = sin a. cos b + cos a. sin b

a= 50° and b=40°, then

sin(50+40) = sin 50° . cos 40° + cos 50° . sin 40°

sin(50+40) = sin 90° = 1 and

sin 50° . cos 40° + cos 50° . sin 40° = 1

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2 years ago

An air transport association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maxim

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Answer:

The 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).

Step-by-step explanation:

The (1 - α)% confidence interval for the population mean, when the population standard deviation is not provided is:

$CI=\bar x \pm t_{\alpha/2, (n-1)}\cdot \frac{s}{\sqrt{n}}$

The sample selected is of size, n = 50.

The critical value of t for 95% confidence level and (n - 1) = 49 degrees of freedom is:

$t_{\alpha/2, (n-1)}=t_{0.05/2, 49}\approx t_{0.025, 60}=2.000$

*Use a t-table.

Compute the sample mean and sample standard deviation as follows:

$\bar x=\frac{1}{n}\sum X=\frac{1}{50}\times [1+5+6+...+10]=6.76\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{49}\times 31.12}=2.552$

Compute the 95% confidence interval estimate of the population mean rating for Miami as follows:

$CI=\bar x \pm t_{\alpha/2, (n-1)}\cdot \frac{s}{\sqrt{n}}$

$=6.76\pm 2.000\times \frac{2.552}{\sqrt{50}}\\\\=6.76\pm 0.722\\\\=(6.038, 7.482)\\\\\approx (6.0, 7.5)$

Thus, the 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).

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