Express p in terms of q if

Mathematics

1 answer:

Irina-Kira [14]3 years ago

8 0

$p=x^2+\dfrac{1}{x^2}\\\\p=\dfrac{x^4}{x^2}+\dfrac{1}{x^2}\\\\p=\dfrac{x^4+1}{x^2}\qquad(*)$

$q=x+\dfrac{1}{x}\\\\q=\dfrac{x^2}{x}+\dfrac{1}{x}\\\\q=\dfrac{x^2+1}{x}\qquad\text{square both sides}\\\\q^2=\left(\dfrac{x^2+1}{x}\right)^2\\\\q^2=\dfrac{(x^2+1)^2}{x^2}\qquad\text{use}\ \ (a+b)^2=a^2+2ab+b^2\\\\q^2=\dfrac{(x^2)^2+2(x^2)(1)+1^2}{x^2}\\\\q^2=\dfrac{x^4+2x^2+1}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+\dfrac{2x^2}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+2\qquad\text{subtract 2 from both sides}\\\\q^2-2=\dfrac{x^4+1}{x^2}\\\\\text{From (*) we have}\\\\\boxed{p=q^2-2}$

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Suppose a batch of metal shafts produced in a manufacturing company have a population standard deviation of 1.3 and a mean diame

lbvjy [14]

Answer:

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

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 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 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.

In this problem, we have that:

$\mu = 208, \sigma = 1.3, n = 60, s = \frac{1.3}{\sqrt{60}} = 0.1678$

What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Lesser than 207.9.

pvalue of Z when X = 207.9. So

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

By the Central Limit Theorem

$Z = \frac{207.9 - 208}{0.1678}$