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jeka57 [31]
3 years ago
13

A gravel company sells gravel by the cubic yard and charges a flat delivery fee. The table shows the total cost C based on the a

mount of gravel in cubic yards g
What is the flat delivery fee (with 0 cubic yards of gravel) in dollars and cents?

Attachment below

Mathematics
1 answer:
Sergio039 [100]3 years ago
6 0

Answer:

<h2>$60.</h2>

Step-by-step explanation:

The cost for 0 cubic yards of gravel refers to the y-intercept of the function.

First, we find the constant ratio of change

r=\frac{284-124}{7-2}=\frac{160}{5}=32

The constant ratio of change is 32, that means the cost per cubic yard of gravel is $32.

Now, we just subtract this ratio twice to find the relations until we have x = 0.

124 -32= 92

So, for 1 cubic yard, the cost is $92.

92-32=60

For 0 cubic yards, the cost is $60, that's the y-intercept or the initial condition.

Therefore, the answer is $60.

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10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

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Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

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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.

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Normally distributed with mean 375 minutes and standard deviation 68 minutes. So \mu = 375, \sigma = 68

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So n = 6, s = \frac{68}{\sqrt{6}} = 27.76

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So n = 6, s = \frac{106}{\sqrt{6}} = 43.27

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

Z = \frac{410 - 523}{43.27}

Z = -2.61

Z = -2.61 has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

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