If it is assumed that the heights of men are normally distributed with a standard deviation of 3.0 inches, how large a sample should be taken to be fairly sure (probability 0.95) that the sample mean does not differ from the true mean (population mean) by more than 0.10 in absolute value? (round your answer up to the nearest whole number.)

3,458 men

Step-by-step explanation:

The confidence interval for a normally distributed parameter can be found by:

Where 'X' is the population mean, 'z' is the z-value for the desire confidence (z=1.960  for 95% confidence), 's' is the standard deviation and 'n' is the sample size.

The population needed to assure that the sample mean does not differ from the true mean by more than 0.10 is:

Rounding up to the next whole unit. The sample size must be at least 3,458 men.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

1. minimum

2. 6400

3. eight

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