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Ilia_Sergeevich [38]
3 years ago
6

For the​ following, indicate whether a confidence interval for a proportion or mean should be constructed to estimate the variab

le of interest. Justify your response. Does chewing your food for a longer period of time reduce​ one's caloric intake of food at​ dinner? A researcher requires a sample of 75 healthy males to chew their food twice as long as they normally do. The researcher then records the calorie consumption at dinner.
Mathematics
2 answers:
Black_prince [1.1K]3 years ago
6 0

Answer:

A confidence interval for a mean

Step-by-step explanation:

anzhelika [568]3 years ago
4 0

Answer:

A confidence interval for a mean should be constructed to estimate the variable of interest.

Step-by-step explanation:

The confidence interval for a mean uses the sample mean to estimate the true population mean.  Since it is a confidence interval, instead of a single number for the mean, the result shows a lower estimate and an upper estimate of the mean.  In this experiment, the confidence interval cannot be for a proportion since the population of healthy males is unknown, but the calculation is being done with a sample of 75 healthy males.

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Alla [95]

Answer:  Choice A

\tan(\alpha)*\cot^2(\alpha)\\\\

============================================================

Explanation:

Recall that \tan(x) = \frac{\sin(x)}{\cos(x)} and \cot(x) = \frac{\cos(x)}{\sin(x)}. The connection between tangent and cotangent is simply involving the reciprocal

From this, we can say,

\tan(\alpha)*\cot^2(\alpha)\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)^2\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\frac{\cos^2(\alpha)}{\sin^2(\alpha)}\\\\\\\frac{\sin(\alpha)*\cos^2(\alpha)}{\cos(\alpha)*\sin^2(\alpha)}\\\\\\\frac{\cos^2(\alpha)}{\cos(\alpha)*\sin(\alpha)}\\\\\\\frac{\cos(\alpha)}{\sin(\alpha)}\\\\

In the second to last step, a pair of sine terms cancel. In the last step, a pair of cosine terms cancel.

All of this shows why \tan(\alpha)*\cot^2(\alpha)\\\\ is identical to \frac{\cos(\alpha)}{\sin(\alpha)}\\\\

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