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3 years ago

Add the following polynomails -3x^2-2xy+y^2,-4x^2-xy+6y^2, x^2+6xy-y^2

Mathematics

1 answer:

Tomtit [17]3 years ago

8 0

Answer:

When we add the given polynomials the result is

$(-3x^2-2xy+y^2)+(-4x^2-xy+6y^2)+(x^2+6xy-y^2)=3(2(y^2-x^2)+xy)$

Step-by-step explanation:

Given polynomials are

$-3x^2-2xy+y^2$ , $-4x^2-xy+6y^2$ , $x^2+6xy-y^2$

Now addding the polynomials we get

$(-3x^2-2xy+y^2)+(-4x^2-xy+6y^2)+(x^2+6xy-y^2)=-3x^2-2xy+y^2+-4x^2-xy+6y^2+x^2+6xy-y^2$

$=-6x^2+3xy+6y^2$ (adding the like terms on RHS)

$=3(-2x^2+xy+2y^2)$ (taking 3 outside on RHS)

$=3(2(-x^2+y^2)+xy)$

Rewritting the above equation we get,

$=3(2(y^2-x^2)+xy)$

Therefore $(-3x^2-2xy+y^2)+(-4x^2-xy+6y^2)+(x^2+6xy-y^2)=3(2(y^2-x^2)+xy)$

When we add the given polynomials the result is

$(-3x^2-2xy+y^2)+(-4x^2-xy+6y^2)+(x^2+6xy-y^2)=3(2(y^2-x^2)+xy)$

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Suppose 45% of the population has a college degree.

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Using the normal distribution, there is a 0.2076 = 20.76% probability that the proportion of persons with a college degree will differ from the population proportion by greater than 3%.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as $np \geq 10$ and $n(1 - p) \geq 10$ .

The proportion estimate and the sample size are given as follows:

p = 0.45, n = 437.

Hence the mean and the standard error are:

$\mu = p = 0.45$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.45(0.55)}{437}} = 0.0238$

The probability that the proportion of persons with a college degree will differ from the population proportion by greater than 3% is 2 multiplied by the p-value of Z when X = 0.45 - 0.03 = 0.42.

Hence:

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

By the Central Limit Theorem:

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