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TEA [102]
2 years ago
7

Interpret the average rate of change of –14/3 that you found previously. What does this mean in terms of the waterslide, from x

= 0 to x = 15?
Mathematics
2 answers:
DaniilM [7]2 years ago
4 0
<span>On average, the slide drops 14 feet for every 3 feet of horizontal distance.
</span><span>On average, the slide drops about 4.7 feet for every 1 foot of horizontal distance.</span>
uranmaximum [27]2 years ago
3 0

Solution: The average rate of change of \frac{-14}{3} represents that the second variable is decrease by 14 if the first variable increase by 3. In terms of water slide the vertical height of slide is decreased at the rate of 4.66 per unit. The vertical height of the slide decrease 70 units from x = 0 to x = 15.

Explanation:

The rate of change shows the change in first variable with respect to change in second variable.

If the rate of chang is given by \frac{-14}{3} it means second variable is decrease by 14 if the first variable increase by 3.

In the context of water slide it shows that the vertical height of slide is decreased by 14 units as we cover the distance of 3 units.

If we cover the distance from x = 0 to x = 15, it means we cover the distance of 15 units.

Let the vertical height of the slide covered from x = 0 to x =15 be y.

It means when distance increased by 15 units the height of slide decreased by y units.

So rate of change = \frac{-y}{15}.

It is given that the rate of change is \frac{-14}{3}.

Equate both equations.

\frac{-y}{15}=\frac{-14}{3}

y=15(\frac{14}{3} )

y=5(14)

y=70

Therefore, the average rate of change of \frac{-14}{3} represents that the second variable is decrease by 14 if the first variable increase by 3. In terms of water slide the vertical height of slide is decreased at the rate of 4.66 per unit. The vertical height of the slide decrease 70 units from x = 0 to x = 15.

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vladimir1956 [14]

Answer:

Cov(X, Y) =0.029.

Step-by-step explanation:

Given that :

The noise in a particular voltage signal has a constant mean of 0.9 V. that is μ = 0.9V ............(1)

Also, the two noise instances sampled τ seconds apart have a bivariate normal distribution with covariance.

0.04e–jτj/10 ............(2)

Having X and Y denoting the noise at times 3 s and 8 s, respectively, the difference of time = 8-3 = 5seconds.

That is, they are 5 seconds apart,

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Thus,

Cov(X, Y), for τ = 5seconds = 0.04e-5/10

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1 year ago
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kiruha [24]

Using the normal distribution and the central limit theorem, it is found that  there is a 0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.

----------------------------------

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

----------------------------------

  • Mean of 700g means that \mu = 700
  • Standard deviation of 21g means that \sigma = 21
  • Sample of 64, thus n = 64
  • <u>For the sampling distribution of the sample mean</u>, the standard deviation is of s = \frac{21}{\sqrt{64}} = \frac{21}{8} = 2.625

The probability of finding a sample mean mass of 695g or below is the p-value of Z when X = 695, thus:

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

By the Central Limit Theorem

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

Z = \frac{695 - 700}{2.625}

Z = -1.905

Z = -1.905 has a p-value of 0.0284.

0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.

A similar problem is given at brainly.com/question/22934264

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