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Paul [167]
3 years ago
15

Prove : (sec θ - tan θ )^2 = 1 - sin θ /1+sin θ

Mathematics
1 answer:
labwork [276]3 years ago
3 0
(sec x - tan x)^2 \\  \\ = sec^2x - 2 sec x tan x + tan^2 x \\  \\ =(1+tan^2 x) - 2 sec x tan x +tan^2 x \\  \\ =1 - 2 sec x tan x + 2 tan^2 x \\  \\ = 1 - 2tan x(sec x - tan x) \\  \\ =1 - \frac{2 sin x}{cos x} (\frac{1-sin x}{cos x}) \\  \\ = 1 - \frac{2 sin x (1-sin x)}{cos^2 x} \\  \\ =1 - \frac{2 sin x (1-sin x)}{1-sin^2 x} \\  \\ =1 - \frac{2 sin x (1-sin x)}{(1-sin x)(1+sin x)} \\  \\ =1-\frac{2 sin x}{1+sin x}
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<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

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In which:

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In this problem, we have a 99% confidence level, hence\alpha = 0.99, z is the value of Z that has a p-value of \frac{1+0.99}{2} = 0.995, so the critical value is z = 2.575.

The estimate and the sample size are given by:

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Then the bounds of the interval are:

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The 99​% confidence interval to estimate the population proportion is: (0.2364, 0.4836).

More can be learned about the z-distribution at brainly.com/question/25890103

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Supplier B would cost more if the company buys 500 bushels of grain per week.

Step-by-step explanation:

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