The table shows the relationship

1 answer:

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The linear relationship which exists between the change in weight and number of days of hibernation in a hedgedog can be modeled using the equation y = - 0.03 + 0

Change in weight per day of hibernation = - 0.03
Change in weight for 115 days = - 3.45 ounces

A linear model can be created using the data in the table given using a regression calculator or excel ;

The linear model obtained in the form y = m + bx is :

y = - 0.03x + 0
y = change in weight
x = number of days of hibernation
Intercept = 0
Slope = -0.03

The slope value of the function gives the change in weight value per number of days of hibernation ; which is -0.03.

Using the regression equation, substitute, the vlaue of x = 115
-0.03(115) + 0 = - 3.45

Therefore, the change in weight after 115 days of hibernation is - 3.45 ounces.

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Assume that the heights of men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. If th

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Answer:

The bottom cutoff heights to be eligible for this experiment is 66.1 inches.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mean of 69.0 inches and a standard deviation of 2.8 inches.

This means that $\mu = 69, \sigma = 2.8$