Expand c(3c - 3) I need help on this

Mathematics

2 answers:

Rudik [331]2 years ago

4 0

$\\ \sf\longmapsto c(3c-3)$

Apply distributive law

a(b-d)=ab-ad

$\\ \sf\longmapsto 3c(c)-3(c)$

$\\ \sf\longmapsto 3c^2-3c$

We can still take common

$\\ \sf\longmapsto 3c(c-1)$

Finger [1]2 years ago

4 0

Answer:

3c² - 3c

Step-by-step explanation:

c(3c - 3)

=> c(3c) - c(3)

=> 3c² - 3c

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A long novel is opened to a random page; the probability that the page is numbered something between 21 and 31.

Harrizon [31]

Using the uniform distribution, considering that the novel has 435 pages, the probability that the page is numbered something between 21 and 31 is:

$p = \frac{10}{435}$

<h3>What is the uniform probability distribution?</h3>

It is a distribution with two bounds, a and b, in which each outcome is equally as likely.

The probability of finding a value between c and d is:

$P(c \leq X \leq d) = \frac{d - c}{b - a}$

Researching the problem on the internet, it is found that the novel has 435 pages, hence a = 0, b = 435, and the probability that the page is numbered something between 21 and 31 is given by:

$P(21 \leq X \leq 31) = \frac{31 - 21}{435 - 0} = \frac{10}{435}$