Why does the fraction 20/25 not belong with 1/20, 2/3 and 5/4?

Mathematics

1 answer:

devlian [24]2 years ago

7 0

I’m not 100% sure but I think it’s because 20/25 can still be simplified further, while the other 3 can’t.

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A continuous probability distribution having a rectangular shape, where the probability is evenly distributed over an interval o

aleksklad [387]

Answer:

A continuous probability distribution having a rectangular shape, where the probability is evenly distributed over an interval of numbers is a(n) __uniform__________ distribution

Step-by-step explanation:

Given that there is a continuous probability distribution having a rectangular shape, where the probability is evenly distributed over an interval of numbers

Since the pdf is rectangular in shape and total probability is one we can say all values in the interval would be equally likely

Say if the interval is (a,b) P(X) = p the same for all places

Since total probability is 1,

we get integral of P(X)=p(b-a) =1

Or p= $\frac{1}{b-a}$

this is nothing but a uniform distribution continuous defined in the interval

A continuous probability distribution having a rectangular shape, where the probability is evenly distributed over an interval of numbers is a(n) __uniform__________ distribution

8 0

2 years ago

Find the common denominator. Click all that applies. 1/4 and 1/2 3/12 and 6/12 3/12 and 6/12 1/4 and 1/2

kozerog [31]

There is a common denominator of 12 among those fractions.

7 0

2 years ago

Plz help this is so hard

Mashutka [201]

1) 5:4 2) 4:5 3) 5:9 4) 2:3 5) 1:4 6)3:2 7) 5:13 8) 6:7 9) 7:5 10) 13:18 11) 18:5 12) 6:13. The others were weird, because it's been forever since I've had to do these.

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3 years ago

Find a power series for the function, centered at c. g(x) = 4x x2 2x − 3 , c = 0

BartSMP [9]

The power series for given function $g(x)=\frac{4x}{(x-1)(x+3)}$ is $g(x)=\sum{_{n=0}^\infty}~x^n(-1+(-\frac{x}{3} )^n)$

For given question,

We have been given a function g(x) = 4x / (x² + 2x - 3)

We need to find a power series for the function, centered at c, for c = 0.

First we factorize the denominator of function g(x), we have:

$\Rightarrow g(x)=\frac{4x}{(x-1)(x+3)}$

We can write g(x) as,

$\Rightarrow g(x)=\frac{1}{x-1}+\frac{3}{x+3}\\\\\Rightarrow g(x)=\frac{-1}{1-x}+\frac{1}{1+\frac{x}{3} }\\\\\Rightarrow g(x)=\frac{-1}{1-x}+\frac{1}{1-(-\frac{x}{3} )}\\$