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drek231 [11]
3 years ago
6

It is important to remember that random coincidences are truly random and _____ correlated

Mathematics
1 answer:
SIZIF [17.4K]3 years ago
4 0
Random coincidences are set of coincidences which have been selected randomly and thus they don't posses any particular relationship at all. This means that they do not have any dependencies whatsoever and thus, the relationship of random event A to random event say B or C or D is purely none. Therefore, whenever A occurs B will not be affected and whenever B or C or D occurs, A will not be affected at all. Therefore random coincidences are truly random and NOT correlated.
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Write the fraction 2/5 as an equivalent with the given denominator 25
OlgaM077 [116]

Answer:

10/25

Step-by-step explanation:

2/5 = x/25

We need to multiply the bottom by 5 to get to 25

5*5 = 25

What we to the bottom, we do the the top

2/5 * 5/5 = 10/25

6 0
3 years ago
Read 2 more answers
Factor of x^3-x^2-24x-36
Basile [38]
The trial and error method is used to find an initial factor:
If we let f(x) = x³ - x² - 24x - 36 and all we have to do is sub' in values of x until
f(x) = 0, we can use this to find an initial factor by the factor theorem:
f(1) = (1)³ - (1)² - 24(1) - 36 = -60
f(2) = (2)³ - (2)² - 24(2) - 36 = -80
f(5) = (5)³ - (5)² - 24(5) - 36 = -56
*** f(6) = (6)³ - (6)² - 24(6) - 36 = 0 ***

f(6) = 0 so (x - 6) is a factor of f(x).
This means that: f(x) = x³ - x² - 24x - 36 = (x - 6)(ax² + bx + c).
To find a,b and c, use long division (or inspection) to divide x³ - x² - 24x - 36 by x - 6.
The other 2 factors of f(x) can then be found by factorizing the
ax² + bx + c quadratic the way you would with any other quadratic (i.e. by quadratic formula, CTS or inspection).
5 0
3 years ago
What would m equal if -6.85+m/4=-11​
BARSIC [14]

Hope this will help u..............

4 0
3 years ago
If a^20=(a^n)^m what are the values for m and n
Alinara [238K]
N = 2 and m=10. You multiply them
7 0
3 years ago
After the fraction x plus 1 all over 2 minus the fraction x plus 2 all over 3 x have been combined using the least common denomi
RoseWind [281]

If I've read this correctly, it looks like this.

\dfrac{(x + 1)}{2 - \dfrac{(x + 2)}{3x}}

If that is correct, then the first step is to put the top part of the denominator over 3x

\dfrac{(x + 1)}{\dfrac{6x - (x + 2)}{3x}} = \dfrac{(x + 1)}{\dfrac{5x -2}{3x}}

The next part is to flip a three tier fraction. I'm afraid I have to show what happens. My latex is not that strong.

What you get is

\dfrac{3x*(x + 1)}{(5x - 2)}

This is just about your final answer. You could write it as

\dfrac{3x^2 + 3x}{(5x - 2)}

6 0
3 years ago
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