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emmainna [20.7K]
3 years ago
13

The Dazzle football team made 5 plays in a row where they gained 3 yards on each play then they had 2 plays in a row where they

lost 12 yards on each play .What is the total change in their position from where they started?
Mathematics
1 answer:
lys-0071 [83]3 years ago
5 0
The total change is position is that they loss 9 yards
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Thinking about the graph of y, the rate of change is zero whenever there is an extremum.

First, differentiate y.

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Next, find the zeros of y' by factoring.

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Now, we substitute these x-values into the original equation to find the coordinates. 

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Lady_Fox [76]

Answer:

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

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 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Proportion of 0.6

This means that p = 0.6

Sample of 46

This means that n = 46

Mean and standard deviation:

\mu = p = 0.6

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6*0.4}{46}} = 0.0722

Probability of obtaining a sample proportion less than 0.5.

p-value of Z when X = 0.5. So

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

By the Central Limit Theorem

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

Z = \frac{0.5 - 0.6}{0.0722}

Z = -1.38

Z = -1.38 has a p-value of 0.0838

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

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The answer is 4√6 4√x
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