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blagie [28]
3 years ago
7

You have collected data about the fasting blood glucose (FBG) level of participants in your study. You are delighted to find tha

t the variable is normally distributed. If the mean FBG is 82 and the standard deviation is 2.8 in what range would you expect to find the FBG of 68% of your study participants
Mathematics
1 answer:
liberstina [14]3 years ago
8 0

Answer:

Between 79.2 and 84.8.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean of 82, standard deviation of 2.8.

In what range would you expect to find the FBG of 68% of your study participants?

By the Empirical Rule, within 1 standard deviation of the mean, so:

82 - 2.8 = 79.2

82 + 2.8 = 84.8

Between 79.2 and 84.8.

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2 Mayra is a caterer and charges $75 per plate plus a
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$11,280

Step-by-step explanation:

Plug in 150 as p into the equation to solve for the cost, C:

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Part A:

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8 0
3 years ago
The sector COB is cut from the circle with center O. The ratio of the area of the sector removed from the whole circle to the ar
mafiozo [28]

Answer:

Ratio = \frac{R^2 - r^2 }{ r^2}

Step-by-step explanation:

Given

See attachment for circles

Required

Ratio of the outer sector to inner sector

The area of a sector is:

Area = \frac{\theta}{360}\pi r^2

For the inner circle

r \to radius

The sector of the inner circle has the following area

A_1 = \frac{\theta}{360}\pi r^2

For the whole circle

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The sector of the outer sector has the following area

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So, the ratio of the outer sector to the inner sector is:

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Cancel out common factor

Ratio = R^2 - r^2 : r^2

Express as fraction

Ratio = \frac{R^2 - r^2 }{ r^2}

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