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White raven [17]
3 years ago
9

What is the hyperbola 25x^2-100y-9y^2-90x-350=0 in standard form?

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
4 0
WE need to complete the square on the x and y terms:-

25x^2 - 100x - 9y^2 - 90y = 350
25(x^2 - 4x +    ) - 9(y^2 + 10x       )  =   350 + 25(  )  - 9(  )
25(x^2 - 4x  + 4 ) - 9(y^2 + 10x  + 25)  = 350  + 25(4)  - 9(25)
25(x - 2)^2 - 9(y + 5)^2 = 

25(x - 2)^2   -  9(y + 5)^2          =  225
----------------    --------------            --------
    225                 225                    225

  ( x  - 2)^2         (y + 5)^2  
  -------------   -   --------------  =    1         Answer
        9                    25                               



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A statue stands on a 625 m2 base. An artist creates a replica using a ratio of 1:10 What is the area of the replica's base?
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<h3>What is the required area?</h3>

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The level of nitrogen oxides (NOX) in a exhaust of cars of a particular model varies normally with mean 0.25 grams per miles and
antoniya [11.8K]

Answer:

a) 15.87% probability that a single car of this model fails to meet the NOX requirement.

b) 2.28% probability that the average NOX level of these cars are above 0.3 g/mi limit

Step-by-step explanation:

We use the normal probability distribution and the central limit theorem to solve this question.

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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 0.25, \sigma = 0.05

a. What is the probability that a single car of this model fails to meet the NOX requirement?

Emissions higher than 0.3, which is 1 subtracted by the pvalue of Z when X = 0.3. So

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

Z = \frac{0.3 - 0.25}{0.05}

Z = 1

Z = 1 has a pvalue of 0.8417.

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15.87% probability that a single car of this model fails to meet the NOX requirement.

b. A company has 4 cars of this model in its fleet. What is the probability that the average NOX level of these cars are above 0.3 g/mi limit?

Now we have n = 4, s = \frac{0.05}{\sqrt{4}} = 0.025

The probability is 1 subtracted by the pvalue of Z when X = 0.3. So

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

By the Central Limit Theorem

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

Z = \frac{0.3 - 0.25}{0.025}

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1 - 0.9772 = 0.0228

2.28% probability that the average NOX level of these cars are above 0.3 g/mi limit

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