All categories

3 years ago

What is the GCF of 36x^2y and 18xy ^2? Please i really need the answers

Mathematics

1 answer:

s2008m [1.1K]3 years ago

8 0

I’m going to show you the answer

You might be interested in

Let f (x) = 1/ x-3and g(x) = x+5. What is the domain of (fºg)(x)?

Natalija [7]

Answer:

I can't find that file. please give it again.

4 0

3 years ago

(36+4) divided by 10 how do I write it in a story problem

BlackZzzverrR [31]

Step-by-step explanation:

you can write it as

the sum of of thirty six(36) and four(4) is divided by Ten(10)

5 0

3 years ago

Read 2 more answers

Practice with lines cut by transversal.<br> What is the value of x?

Ghella [55]

X should equal 70 I think.

8 0

3 years ago

Read 2 more answers

PLEASE HELP!!! Help would be really appreciated here!!! <br> Find M angle V

Blizzard [7]

Answer is 16 ° of angels

4 0

3 years ago

Find the cdf F(x) associated with each of the following probability density functions. Sketch the graphs of f(x) and F(x).

wolverine [178]

Answer:

See steps below

Step-by-step explanation:

$\bf f(x)=3(1-x)^2\;(0$

$\bf F(x)=\int_{0}^{x}f(t)dt=\int_{0}^{x}3(1-t)^2dt=3\int_{0}^{x}(1-t)^2=1-(1-x)^3$

The cdf associated with f is

$\bf \boxed{F(x)=1-(1-x)^3}$ for 0<x<1

<h3>See picture 1 </h3>

The median is a point x such that

F(x) = ½

so, the median is

$\bf 1-(1-x)^3=1/2\rightarrow (1-x)^3=1/2\rightarrow \boxed{x=1-\sqrt[3]{2}}$

The 25th percentile equals the 1st quartile and is a point x such

F(x) = ¼

and the 25th percentile is

$\bf 1-(1-x)^3=1/4\rightarrow (1-x)^3=3/4\rightarrow \boxed{x=1-\sqrt[3]{3/4}}$

$\bf f(x)=\frac{1}{x^2}\;(1$

$\bf F(x)=\int_{1}^{x}f(t)dt=\int_{1}^{x}\frac{dt}{t^2}=1-\frac{1}{x}$

The cdf associated with f is

$\bf \boxed{F(x)=1-\frac{1}{x}}$ for x>1

<h3>See picture 2 </h3>

The median is

$\bf 1-\frac{1}{x}=\frac{1}{2}\rightarrow \boxed{x=2}$

The 25th percentile is

$\bf 1-\frac{1}{x}=\frac{1}{4}\rightarrow \boxed{x=4/3}$

f(x) = 1/3 for 0<x<1 or 2<x<4

$\bf F(x)=\int_{0}^{x}\frac{dt}{3}=\frac{x}{3}\;(0$

$\bf F(x)=\frac{1}{3}+\int_{2}^{x}\frac{dt}{3}=\frac{1}{3}+\frac{x-2}{3}=\frac{x-1}{3}\;(2$

The cdf associated with f is

$\bf F(x)=\frac{x}{3}$ for 0<x<1

$\bf F(x)=\frac{x-1}{3}$ for 2<x<4

<h3>See picture 3 </h3>

The median is

$\bf \frac{x-1}{3}=1/2\rightarrow x=1+3/2\rightarrow \boxed{x=5/2}$

The 25th percentile is

$\bf \frac{x}{3}=1/4\rightarrow \boxed{x=3/4}$

4 0

4 years ago