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Ahat [919]
3 years ago
10

Explain whether the points (−13,4), (−7,3), (−1,2), (5,1), (11,0), (17,−1) represent the set of all the solutions for the equati

on y=−16x+116.
Yes, because all the solutions of y=−16x+116 need to be integer values for x and y.

Yes, because all of the points that are listed are on the line y=−16x+116.

No, because the set of all solutions of y=−16x+116 is represented by the line of the equation.

No, because the point (17,−1) is not on the line that represents the equation y=−16x+116.
Mathematics
1 answer:
ivann1987 [24]3 years ago
8 0

Answer:

No, because the set of all solutions of y=−16x+116 is represented by the line of the equation.

Step-by-step explanation:

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Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and
TiliK225 [7]

Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central limit Theorem

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}}

In this problem, we have that:

\mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33

What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

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

By the Central Limit Theorem

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

2.33 = \frac{X - 117}{4.33}

X - 117 = 2.33*4.33

X = 127.1

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

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Answer:

Step-by-step explanation:

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Answer:

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21 rounded to the nearest 10 is 20

so I don't really know but I guess you should put the x as 20 for sure UwU

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