(5x^5) - (80x^3) factoring

Mathematics

1 answer:

elena55 [62]3 years ago

4 0

Answer:

Factoring the term $(5x^5) - (80x^3)$ we get $\mathbf{5x^3(x-4)(x+4)}$

Step-by-step explanation:

We need to factor the term $(5x^5) - (80x^3)$

First we can see that $5x^3$ is common in both terms

So, taking $5x^3$ common:

$(5x^5) - (80x^3)\\=5x^3(x^2-16)$

We can write $x^2-16$ as $(x)^2-(4)^2$

$=5x^3((x)^2-(4)^2)$

Now we can solve $(x)^2-(4)^2$ using the formula: $a^2-b^2=(a+b)(a-b)$

We can write:

$=5x^3((x)^2-(4)^2)\\=5x^3(x-4)(x+4)$

So, factoring the term $(5x^5) - (80x^3)$ we get $\mathbf{5x^3(x-4)(x+4)}$

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In a randomly selected sample of 1169 men ages 35–44, the mean total cholesterol level was 210 milligrams per deciliter with a s

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Answer:

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Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 210, \sigma = 38.6$