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3 years ago

CosA + cosB + cosC = 1 + 4sinA/2 sinB/2 sinC/2

1 answer:

6 0

$\cos(A) + \cos( B ) + \cos(C) \\ = 2 \cos( \frac{A + B}{2} ) \cos( \frac{ A- B}{2} ) + 1 - 2 { \sin }^{2} \frac{C}{2} \\ = 1 + 2\cos( \frac{\pi - C}{2} ) \cos( \frac{A - B}{2} ) - 2 { \sin }^{2} \frac{C}{2} \\ = 1 + 2 \sin \frac{C}{2} \cos( \frac{A - B}{2} ) - 2 { \sin }^{2} \frac{C}{2} \\ = 1 + 2 \sin \frac{C}{2} [ \cos( \frac{A - B}{2} ) - \sin \frac{C}{2}] \\ = 1 + 2 \sin \frac{C}{2} [ \cos( \frac{A - B}{2} ) - \cos( \frac{A + B}{2} )] \\ = 1 + 2 \sin\frac{C}{2} \times 2 \sin( \frac{ \frac{A + B}{2} +\frac{A - B}{2} }{2} ) \sin( \frac{ \frac{A + B}{2} - \frac{A - B}{2} }{2} ) \\ = 1 + 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$

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$\mu_p -\sigma_p = 0.74-0.0219=0.718$

$\mu_p +\sigma_p = 0.74+0.0219=0.762$

68% of the rates are expected to be betwen 0.718 and 0.762

$\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696$

$\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784$

95% of the rates are expected to be betwen 0.696 and 0.784

$\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674$

$\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806$

99.7% of the rates are expected to be betwen 0.674 and 0.806

Step-by-step explanation:

Check for conditions

For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:

a) Independence : we assume that the random sample of 400 students each student is independent from the other.

b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.

c) np= 400*0.74= 296>10

n(1-p) = 400*(1-0.74)=104>10

So then we have all the conditions satisfied.

Solution to the problem

For this case we know that the distribution for the population proportion is given by:

$p \sim N(p, \sqrt{\frac{p(1-p)}{n}})$

So then:

$\mu_p = 0.74$

$\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219$

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

$\mu_p -\sigma_p = 0.74-0.0219=0.718$

$\mu_p +\sigma_p = 0.74+0.0219=0.762$

68% of the rates are expected to be betwen 0.718 and 0.762

$\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696$

$\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784$

95% of the rates are expected to be betwen 0.696 and 0.784

$\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674$

$\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806$

99.7% of the rates are expected to be betwen 0.674 and 0.806

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3 years ago