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Alchen [17]
2 years ago
7

A skateboarder starts from rest and maintains a constant forward acceleration of 0.50 meter per second squared for 8.4 s. what i

s the rider’s displacement during this time?
Mathematics
2 answers:
myrzilka [38]2 years ago
7 0

Answer-

The rider’s displacement during this time is 17.01 m

Solution-

As we know,

s= ut + \frac{1}{2}at^{2}

Where s = displacement,

           u= initial velocity = 0, as the rider starts from rest

           a= acceleration = 0.5 m/s²

           t= time taken = 8.4 s

Putting the values,

s = (0)(8.4) + \frac{1}{2} (0.5)(8.4)^{2} = 17.64 \ m

NeX [460]2 years ago
7 0

Answer:


Step-by-step explanation:

a=0.50

t=8.4s

initial velocity=0

0+1/2*0.50*8.4^2=17.64m

17.6


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  D.  StartFraction negative 1 + StartRoot 3 EndRoot Over 2 EndFraction

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A manager at a local manufacturing company has been monitoring the output of one of the machines used to manufacture chromium sh
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Answer:

0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

Step-by-step explanation:

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

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 estabilishes 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.

Normally distributed with a mean of 118 centimeters and a standard deviation of 8 centimeters.

This means that \mu = 118, \sigma = 8

Sample of 16 shells

This means that n = 16, s = \frac{8}{\sqrt{16}} = 2

What is the probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly?"

This is the pvalue of Z when X = 120 subtracted by the pvalue of Z when X = 116.

X = 120

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

By the Central Limit Theorem

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

Z = \frac{120 - 118}{2}

Z = 1

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X = 116

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

Z = \frac{116 - 118}{2}

Z = -1

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0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

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A plane is flying at 1000 meters above the ground and fires a laser 60 degrees of of straight down. what is the distance the lig
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Answer:

Step-by-step explanation:

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